下面是线段树的模板，只需要简单地修改就可以将其运用到题目当中

// 线段树的信息
const int N = 2e5 + 10,mod = 1e9 + 7;
int a[N];
struct info // 存储线段树的值
{
    int size;
    int num;
};

struct tag // 存储线段树的懒标记
{
    int add;
    int mul;
};

struct Node // 线段树
{
    info val;
    tag lazy;
}tr[N << 2];
int st_size; // 线段树的总区间大小

// 线段树的具体操作
info operator + (const info &l,const info &r) // pushup的操作
{
    info c;
    c.size = l.size + r.size;
    c.num = (1ll * l.num + r.num) % mod;
    return c;
}

info operator + (const info &v,const tag &t) // pushdown时，对子节点info的操作
{
    info c;
    c.size = v.size;
    c.num = (1ll * v.num * t.mul + 1ll * v.size * t.add) % mod;
    return c;
}

tag operator + (const tag &ts,const tag &tp) // pushdown时，对子节点tag的操作
{
    tag c;
    c.add = (1ll * ts.add * tp.mul + tp.add) % mod;
    c.mul = 1ll * ts.mul * tp.mul % mod;
    return c;
}

void pushup(int u)
{
    tr[u].val = tr[u << 1].val + tr[u << 1 | 1].val;
}

void pushdown(int u)
{
    Node &t = tr[u],&l = tr[u << 1],&r = tr[u << 1 | 1];
    l.val = l.val + t.lazy,l.lazy = l.lazy + t.lazy;
    r.val = r.val + t.lazy,r.lazy = r.lazy + t.lazy;
    t.lazy = {0,1};
}

void build(int u = 1, int l = 1, int r = st_size)
{
    if (l == r)
    {
        tr[u] = {r - l + 1, a[l], 0, 1};
        return;
    }
    tr[u] = {r - l + 1, 0, 0, 1};
    int mid = l + r >> 1;
    build(u << 1, l, mid), build(u << 1 | 1, mid + 1, r);
    pushup(u);
}

void modify(int u, int l, int r,tag c, int pl = 1, int pr = st_size)
{
    if (l <= pl && pr <= r)
    {
        tr[u].val = tr[u].val + c;
        tr[u].lazy = tr[u].lazy + c;
        return;
    }
    pushdown(u);
    int mid = pl + pr >> 1;
    if (l <= mid)
        modify(u << 1, l, r, c, pl, mid);
    if (r > mid)
        modify(u << 1 | 1, l, r, c, mid + 1, pr);
    pushup(u);
}

int query(int u, int l, int r, int pl = 1, int pr = st_size)
{
    if (l <= pl && pr <= r)
        return tr[u].val.num;
    int mid = pl + pr >> 1;
    pushdown(u);
    int res = 0;
    if (l <= mid)
        res = (1ll * res + query(u << 1, l, r, pl, mid)) % mod;
    if (r > mid)
        res = (1ll * res + query(u << 1 | 1, l, r, mid + 1, pr)) % mod;
    return res;
}