/*
 *
 * 思路，将点拓展为三维(x, y, z) z为1表示是查询中的点，z为0表示原来图上就已经有的点，
 * 每一个查询都对应的是二维差分问题，关键问题就是怎么对一个坐标点(x, y), 求xi <= x
 * 且 yi <= y的所有点的权值和，若加上z这一维度，刚好是对z=1的点(x, y, z), 求所有的
 * xi <= x and y1 <= y and zi = 0 的所有点的权值和，这样就可以转成CDQ分治来做了
 *
 */

#include <stdio.h>
#include <string.h>
#include <map>
#include <algorithm>
using namespace std;



const int MAX_NODE_NUM = 1000000;

// 三元组,表示三个维度属性值
struct Node {
    int a, b;                   // a, b对应 x, y坐标
    int c;                      // c数值为0， 1， 0表示原来图上存在的点，1表示询问中涉及到的点
    long long sum;              // 在当前点左下角的所有c为0的点的权值和
    long long wei;              // 点权值

    bool operator < (const Node& other) const {
        if (a != other.a) return a < other.a;
        if (b != other.b) return b < other.b;
        return c < other.c;
    }

    bool operator == (const Node& other) {
        return a == other.a && b == other.b && c == other.c;
    }

} buf[1000000], merge_buf[1000000], q[1000000];     // buf是输入数据的缓存，也是结果缓存，merge_buf是归并排序时候使用的额外缓存



void merge_sort(int ll, int rr) {
    if (ll == rr) {
        return;
    }

    int mid = (ll + rr) >> 1;
    merge_sort(ll, mid);
    merge_sort(mid+1, rr);

    int ii = ll - 1;
    int pos = -1;
    long long tot = 0;
    for (int jj = mid + 1; jj <= rr; jj++) {
        while (ii+1 <= mid && buf[ii+1].b <= buf[jj].b) {
            ii += 1;

            if (buf[ii].c == 0) {
                tot += buf[ii].wei;
            }

            merge_buf[++pos] = buf[ii];
        }

        if (buf[jj].c == 1) {
            buf[jj].sum += tot;
        }
        merge_buf[++pos] = buf[jj];
    }

    while (ii+1 <= mid) {
        ii += 1;
        merge_buf[++pos] = buf[ii];
    }

    // 子区间合并
    for (int i = ll; i <= rr; i++) {
        buf[i] = merge_buf[i-ll];
    }
}

// n是输入的三元组数组的长度
void cdq(int n) {
    sort(buf, buf + n);
    merge_sort(0, n-1);
}


int ans[1000000];
int main() {
    // freopen("/Users/grh/Programming/CLionProjects/SimpleTest/input.txt", "r", stdin);

    int n, m;
    int a, b, w;
    scanf("%d %d", &n, &m);

    int pos = 0;
    for (int i = 0; i < n; i++) {
        scanf("%d %d %lld", &(buf[pos].a), &(buf[pos].b), &(buf[pos].wei));
        buf[pos].c = 0;
        buf[pos].sum = 0;
        pos += 1;
    }

    int x1, y1, x2, y2;
    int q_pos = 0;
    for (int i = 0; i < m; i++) {
        scanf("%d %d %d %d", &x1, &y1, &x2, &y2);
        if (x1 > x2) {
            swap(x1, x2);
        }

        if (y1 > y2) {
            swap(y1, y2);
        }

        buf[pos].a = x2, buf[pos].b = y2, buf[pos].c = 1;
        buf[pos].wei = 0, buf[pos].sum = 0;
        q[q_pos++] = buf[pos];
        pos += 1;

        buf[pos].a = x1-1, buf[pos].b = y2, buf[pos].c = 1;
        buf[pos].wei = 0, buf[pos].sum = 0;
        q[q_pos++] = buf[pos];
        pos += 1;

        buf[pos].a = x2, buf[pos].b = y1-1, buf[pos].c = 1;
        buf[pos].wei = 0, buf[pos].sum = 0;
        q[q_pos++] = buf[pos];
        pos += 1;

        buf[pos].a = x1-1, buf[pos].b = y1-1, buf[pos].c = 1;
        buf[pos].wei = 0, buf[pos].sum = 0;
        q[q_pos++] = buf[pos];
        pos += 1;
    }

    cdq(pos);

    map<Node, long long> nodes;
    for (int i = 0; i < pos; i++) {
        if (buf[i].c == 1) {
            nodes[buf[i]] = buf[i].sum;
        }
    }

    // 用矩形的四个关键点来进行差分计算
    for (int i = 0; i < q_pos >> 2; i++) {
        long long tot = 0;
        tot += nodes[q[i * 4]];
        tot -= nodes[q[i * 4 + 1]];
        tot -= nodes[q[i * 4 + 2]];
        tot += nodes[q[i * 4 + 3]];
        printf("%lld\n", tot);
    }

    return 0;
}