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;
const int kmod = 1000000007;
const int N = 100010;
int n, m, a[N], val[N << 2], c[6][6];
long long f[6][N << 2], s[6][N], p[N][6];
void Push_up(int v, int l, int r, int rt) {
  for (int i = 0; i < 6; ++i)
    f[i][rt] = (s[i][r] - s[i][l - 1] + kmod) % kmod * v % kmod;
  val[rt] = v;
}
void update(int v, int L, int R, int l, int r, int rt) {
  if (l == L && r == R) {
    Push_up(v, l, r, rt);
    return;
  }
  int mid = (l + r) / 2;
  if (val[rt] >= 0) {
    Push_up(val[rt], l, mid, rt << 1);
    Push_up(val[rt], mid + 1, r, rt << 1 | 1);
    val[rt] = -1;
  }
  if (L <= mid) update(v, L, min(mid, R), l, mid, rt << 1);
  if (mid < R) update(v, max(L, mid + 1), R, mid + 1, r, rt << 1 | 1);
  for (int i = 0; i < 6; ++i)
    f[i][rt] = (f[i][rt << 1] + f[i][rt << 1 | 1]) % kmod;
}
int get(int L, int R, int z, int o, int l, int r, int rt) {
  if (l == L && R == r) {
    long long res = 0;
    for (int i = 0; i <= z; ++i)
      res += f[z - i][rt] * p[o][i] % kmod * c[z][i] % kmod;
    res %= kmod;
    if (res < 0) res += kmod;
    return int(res);
  }
  int mid = (l + r) / 2, res = 0;
  if (val[rt] >= 0) {
    Push_up(val[rt], l, mid, rt << 1);
    Push_up(val[rt], mid + 1, r, rt << 1 | 1);
    val[rt] = -1;
  }
  if (L <= mid) res += get(L, min(R, mid), z, o, l, mid, rt << 1);
  if (mid < R) res += get(max(L, mid + 1), R, z, o, mid + 1, r, rt << 1 | 1);
  return res % kmod;
}
int main() {
  memset(val, -1, sizeof val);
  for (int i = 0; i <= 100000; ++i) {
    p[i][0] = 1;
    for (int j = 1; j < 6; ++j) p[i][j] = p[i][j - 1] * i % kmod;
  }
  for (int i = 0; i < 6; ++i)
    for (int j = 0; j <= i; ++j) {
      c[i][j] = j ? abs(c[i - 1][j]) + abs(c[i - 1][j - 1]) : 1;
      if (j % 2) c[i][j] = -c[i][j];
    }
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
    for (int j = 0; j < 6; ++j) s[j][i] = (s[j][i - 1] + p[i][j]) % kmod;
    update(a[i], i, i, 1, n, 1);
  }
  getchar();
  while (m--) {
    char ch;
    int x, y, z;
    scanf("%c %d %d %d", &ch, &x, &y, &z);
    getchar();
    if (ch == '=')
      update(z, x, y, 1, n, 1);
    else
      printf("%d\n", get(x, y, z, x - 1, 1, n, 1));
  }
  return 0;
}