Problem: You've got an array, consisting of n integers: a1, a2, ..., an. Your task is to quickly run the queries of two types:

  1. Assign value x to all elements from l to r inclusive. After such query the values of the elements of array al, al + 1, ..., ar become equal to x.
  2. Calculate and print sum <image>, where k doesn't exceed 5. As the value of the sum can be rather large, you should print it modulo 1000000007 (109 + 7).

Input

The first line contains two integers n and m (1 ≤ n, m ≤ 105), showing, how many numbers are in the array and the number of queries, correspondingly. The second line contains n integers: a1, a2, ..., an (0 ≤ ai ≤ 109) — the initial values of the array elements.

Then m queries follow, one per line:

  1. The assign query has the following format: "<image>", (1 ≤ l ≤ r ≤ n; 0 ≤ x ≤ 109).
  2. The query to calculate the sum has the following format: "<image>", (1 ≤ l ≤ r ≤ n; 0 ≤ k ≤ 5).



All numbers in the input are integers.

Output

For each query to calculate the sum print an integer — the required sum modulo 1000000007 (109 + 7).

Examples

Input

4 5
5 10 2 1
? 1 2 1
= 2 2 0
? 2 4 3
= 1 4 1
? 1 4 5


Output

25
43
1300


Input

3 1
1000000000 1000000000 1000000000
? 1 3 0


Output

999999986

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
const int N = 100005, mod = 1000000007;
int n, m, pws[N][6], sum[N * 4][6], cov[N * 4], C[6][6];
void pushup(int u) {
  for (int i = 0; i <= 5; i++)
    sum[u][i] = (sum[u << 1][i] + sum[u << 1 | 1][i]) % mod;
}
void make(int u, int l, int r, int v) {
  cov[u] = v;
  for (int i = 0; i <= 5; i++)
    sum[u][i] = 1ll * (pws[r][i] - pws[l - 1][i] + mod) * v % mod;
}
void build(int u, int l, int r) {
  cov[u] = -1;
  if (l == r) {
    scanf("%d", sum[u]);
    make(u, l, r, sum[u][0]);
    return;
  }
  int mid = l + r >> 1;
  build(u << 1, l, mid);
  build(u << 1 | 1, mid + 1, r);
  pushup(u);
}
void pushdown(int u, int l, int r, int mid) {
  if (cov[u] != -1) {
    make(u << 1, l, mid, cov[u]);
    make(u << 1 | 1, mid + 1, r, cov[u]);
    cov[u] = -1;
  }
}
int query(int u, int l, int r, int ql, int qr, int k) {
  if (l >= ql && r <= qr) return sum[u][k];
  int mid = l + r >> 1, sum = 0;
  pushdown(u, l, r, mid);
  if (ql <= mid) sum = query(u << 1, l, mid, ql, qr, k);
  if (qr > mid) sum = (sum + query(u << 1 | 1, mid + 1, r, ql, qr, k)) % mod;
  return sum;
}
void modify(int u, int l, int r, int ql, int qr, int v) {
  if (l >= ql && r <= qr) {
    make(u, l, r, v);
    return;
  }
  int mid = l + r >> 1;
  pushdown(u, l, r, mid);
  if (ql <= mid) modify(u << 1, l, mid, ql, qr, v);
  if (qr > mid) modify(u << 1 | 1, mid + 1, r, ql, qr, v);
  pushup(u);
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) {
    int w = 1;
    for (int j = 0; j <= 5; j++) {
      pws[i][j] = (pws[i - 1][j] + w) % mod;
      w = 1ll * w * i % mod;
    }
  }
  for (int i = 0; i <= 5; i++)
    for (int j = C[i][0] = 1; j <= i; j++)
      C[i][j] = C[i - 1][j] + C[i - 1][j - 1];
  build(1, 1, n);
  while (m--) {
    char op[5];
    int l, r, x;
    scanf("%s%d%d%d", op, &l, &r, &x);
    if (op[0] == '=')
      modify(1, 1, n, l, r, x);
    else {
      int ans = 0, w = 1;
      for (int j = x; j >= 0; j--) {
        ans = (ans + 1ll * query(1, 1, n, l, r, j) * C[x][j] % mod * w) % mod;
        w = w * (mod + 1ll - l) % mod;
      }
      printf("%d\n", ans);
    }
  }
}