#include <iostream>
using namespace std;
const int N = 100005;
int n, q, k, a[N];
int main() {
    scanf("%d%d", &n, &q);
    for (int i = 0; i < n; i++)  
    scanf("%d", &a[i]);
    while (q--) {
        scanf("%d", &k);
        int l = 0, r = n - 1;
        while (l < r) {
            int mid = l + r >> 1;
            if (a[mid] < k)  l = mid + 1;//当a[mid]<k时，即左半边都小于k，而我们要找大于等于k的第一个数，所以令左半边的l从mid的右边找到答案
            else  r = mid;////当a[mid]>k时，即右半边都大于k，而我们要找大于等于k的第一个数，所以令右半边r从mid的左边找到答案
        }
        if (a[l] != k) {
            printf("-1 -1\n");
            continue;
        }
        int l1 = l, r1 = n - 1;
        while (l1  < r1) {
            int mid = (l1 + r1 + 1) >> 1;//整数二分边界问题，先简单理解成如果是l1等于mid则需要+1
            if (a[mid] <= k)  l1 = mid;
            else    r1 = mid-1;
        }
        printf("%d %d\n", l, l1);
    }
    return 0;
}