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>
using namespace std;
int c[100005][6][6];
const int modo = 1000000007;
int sum[100005][6];
int val[1 << 18][6];
int a[100005];
int tag[1 << 18];
void updates(int num, int delta) {
  register int i;
  for (i = 0; i <= 5; i++) {
    val[num][i] = val[num * 2 + 1][i];
    int j;
    for (j = 0; j <= i; j++) {
      val[num][i] =
          (val[num][i] + (long long)val[num * 2 + 2][j] * c[delta][i][j]) %
          modo;
    }
  }
}
int sums;
int k;
void update(int num, int delta) {
  int j;
  for (j = 0; j <= k; j++) {
    sums = (sums + (long long)val[num][j] * c[delta][k][j]) % modo;
  }
}
void push_down(int num, int l, int mid, int r) {
  if (tag[num] == -1) return;
  int i;
  for (i = 0; i <= 5; i++) {
    val[num * 2 + 1][i] = (long long)sum[mid - l][i] * tag[num] % modo;
    val[num * 2 + 2][i] = (long long)sum[r - mid][i] * tag[num] % modo;
  }
  tag[num * 2 + 1] = tag[num];
  tag[num * 2 + 2] = tag[num];
  tag[num] = -1;
}
void build_tree(int num, int l, int r) {
  tag[num] = -1;
  if (l == r - 1) {
    int i;
    for (i = 0; i <= 5; i++) {
      val[num][i] = a[l];
    }
    return;
  }
  int mid = (l + r) / 2;
  build_tree(num * 2 + 1, l, mid);
  build_tree(num * 2 + 2, mid, r);
  updates(num, mid - l);
}
void change(int num, int l, int r, int l0, int r0, int vals) {
  if ((l0 <= l) && (r <= r0)) {
    tag[num] = vals;
    int i;
    for (i = 0; i <= 5; i++) {
      val[num][i] = (long long)sum[r - l][i] * vals % modo;
    }
    return;
  }
  int mid = (l + r) / 2;
  push_down(num, l, mid, r);
  if (l0 < mid) change(num * 2 + 1, l, mid, l0, r0, vals);
  if (mid < r0) change(num * 2 + 2, mid, r, l0, r0, vals);
  updates(num, mid - l);
}
void query(int num, int l, int r, int l0, int r0) {
  if ((l0 <= l) && (r <= r0)) {
    update(num, l - l0);
    return;
  }
  int mid = (l + r) / 2;
  push_down(num, l, mid, r);
  if (l0 < mid) query(num * 2 + 1, l, mid, l0, r0);
  if (mid < r0) query(num * 2 + 2, mid, r, l0, r0);
}
int main() {
  int i;
  for (i = 0; i <= 100000; i++) {
    int j;
    for (j = 0; j <= 5; j++) {
      int k;
      c[i][j][0] = 1;
      for (k = 1; k <= j; k++) {
        c[i][j][k] = c[i][j - 1][k] + c[i][j - 1][k - 1];
        if (c[i][j][k] >= modo) c[i][j][k] -= modo;
      }
    }
    for (j = 0; j <= 5; j++) {
      int k;
      c[i][j][0] = 1;
      int p = 1;
      for (k = j; k >= 0; k--) {
        c[i][j][k] = (long long)c[i][j][k] * p % modo;
        p = (long long)p * i % modo;
      }
    }
  }
  for (i = 1; i <= 100000; i++) {
    int j;
    int p = 1;
    for (j = 0; j <= 5; j++) {
      sum[i][j] = sum[i - 1][j] + p;
      if (sum[i][j] >= modo) sum[i][j] -= modo;
      p = (long long)p * i % modo;
    }
  }
  int n;
  scanf("%d", &n);
  int q;
  scanf("%d", &q);
  for (i = 0; i < n; i++) {
    scanf("%d", &a[i]);
  }
  build_tree(0, 0, n);
  for (i = 0; i < q; i++) {
    static char b[15];
    scanf("%s", b);
    int l, r, z;
    scanf("%d%d%d", &l, &r, &z);
    l--;
    if (b[0] == '=') {
      change(0, 0, n, l, r, z);
    } else {
      sums = 0;
      k = z;
      query(0, 0, n, l, r);
      printf("%d\n", sums);
    }
  }
  return 0;
}