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 mod = 1000000007;
int n, m;
int a[100001];
long long cum[6][100001];
long long tree[6][1 << 18];
long long lazy[6][1 << 18];
inline void build(int l, int r, int k, int i = 1) {
  if (l == r) {
    tree[k][i] = (a[l] * (cum[k][l] - cum[k][l - 1] + mod)) % mod;
    lazy[k][i] = -1;
    return;
  }
  int mid = (l + r) >> 1, li = i << 1, ri = li + 1;
  build(l, mid, k, li);
  build(mid + 1, r, k, ri);
  tree[k][i] = (tree[k][li] + tree[k][ri]) % mod;
  lazy[k][i] = -1;
}
inline void update(int l, int r, int s, int t, int x, int k, int i = 1) {
  if (t < l || r < s) return;
  if (l >= s && r <= t) {
    tree[k][i] = ((cum[k][r] - cum[k][l - 1] + mod) * x) % mod;
    lazy[k][i] = x;
    return;
  }
  int mid = (l + r) >> 1, li = i << 1, ri = li + 1;
  if (lazy[k][i] != -1) {
    tree[k][li] = (lazy[k][i] * (cum[k][mid] - cum[k][l - 1] + mod)) % mod;
    lazy[k][li] = lazy[k][i];
    tree[k][ri] = (lazy[k][i] * (cum[k][r] - cum[k][mid] + mod)) % mod;
    lazy[k][ri] = lazy[k][i];
    lazy[k][i] = -1;
  }
  update(l, mid, s, t, x, k, li);
  update(mid + 1, r, s, t, x, k, ri);
  tree[k][i] = (tree[k][li] + tree[k][ri]) % mod;
}
inline long long query(int l, int r, int s, int t, int k, int i = 1) {
  if (t < l || r < s) return 0;
  if (l >= s && r <= t) {
    return tree[k][i];
  }
  int mid = (l + r) >> 1, li = i << 1, ri = li + 1;
  if (lazy[k][i] != -1) {
    tree[k][li] = (lazy[k][i] * (cum[k][mid] - cum[k][l - 1] + mod)) % mod;
    lazy[k][li] = lazy[k][i];
    tree[k][ri] = (lazy[k][i] * (cum[k][r] - cum[k][mid] + mod)) % mod;
    lazy[k][ri] = lazy[k][i];
    lazy[k][i] = -1;
  }
  return (query(l, mid, s, t, k, li) + query(mid + 1, r, s, t, k, ri)) % mod;
}
int C[6][6];
inline int binomial(int l, int r, int k) {
  long long ans = 0;
  long long L = 1;
  for (int i = 0; i <= k; ++i) {
    ans = (ans + ((query(1, n, l, r, k - i) * L) % mod * C[k][i]) % mod) % mod;
    L = (L * (mod - (l - 1))) % mod;
  }
  return (int)ans;
}
int main() {
  for (int i = 0; i < 6; ++i) {
    for (int j = 0; j <= i; ++j) {
      if (i == 0 || j == 0 || i == j)
        C[i][j] = 1;
      else
        C[i][j] = C[i - 1][j] + C[i - 1][j - 1];
    }
  }
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= n; ++i) scanf("%d", a + i);
  for (int i = 1; i <= n; ++i) {
    long long p = 1;
    for (int k = 0; k < 6; ++k) {
      cum[k][i] = (cum[k][i - 1] + p + mod) % mod;
      p = (p * i) % mod;
    }
  }
  for (int k = 0; k < 6; ++k) build(1, n, k);
  while (m--) {
    char op;
    int l, r, x;
    scanf(" %c %d %d %d", &op, &l, &r, &x);
    if (op == '?') {
      printf("%d\n", binomial(l, r, x));
    } else {
      for (int k = 0; k < 6; ++k) update(1, n, l, r, x, k);
    }
  }
  return 0;
}