Problem: At the children's day, the child came to Picks's house, and messed his house up. Picks was angry at him. A lot of important things were lost, in particular the favorite sequence of Picks.

Fortunately, Picks remembers how to repair the sequence. Initially he should create an integer array a[1], a[2], ..., a[n]. Then he should perform a sequence of m operations. An operation can be one of the following:

  1. Print operation l, r. Picks should write down the value of <image>. 
  2. Modulo operation l, r, x. Picks should perform assignment a[i] = a[i] mod x for each i (l ≤ i ≤ r). 
  3. Set operation k, x. Picks should set the value of a[k] to x (in other words perform an assignment a[k] = x). 



Can you help Picks to perform the whole sequence of operations?

Input

The first line of input contains two integer: n, m (1 ≤ n, m ≤ 105). The second line contains n integers, separated by space: a[1], a[2], ..., a[n] (1 ≤ a[i] ≤ 109) — initial value of array elements.

Each of the next m lines begins with a number type <image>. 

  * If type = 1, there will be two integers more in the line: l, r (1 ≤ l ≤ r ≤ n), which correspond the operation 1. 
  * If type = 2, there will be three integers more in the line: l, r, x (1 ≤ l ≤ r ≤ n; 1 ≤ x ≤ 109), which correspond the operation 2. 
  * If type = 3, there will be two integers more in the line: k, x (1 ≤ k ≤ n; 1 ≤ x ≤ 109), which correspond the operation 3. 

Output

For each operation 1, please print a line containing the answer. Notice that the answer may exceed the 32-bit integer.

Examples

Input

5 5
1 2 3 4 5
2 3 5 4
3 3 5
1 2 5
2 1 3 3
1 1 3


Output

8
5


Input

10 10
6 9 6 7 6 1 10 10 9 5
1 3 9
2 7 10 9
2 5 10 8
1 4 7
3 3 7
2 7 9 9
1 2 4
1 6 6
1 5 9
3 1 10


Output

49
15
23
1
9

Note

Consider the first testcase:

  * At first, a = {1, 2, 3, 4, 5}. 
  * After operation 1, a = {1, 2, 3, 0, 1}. 
  * After operation 2, a = {1, 2, 5, 0, 1}. 
  * At operation 3, 2 + 5 + 0 + 1 = 8. 
  * After operation 4, a = {1, 2, 2, 0, 1}. 
  * At operation 5, 1 + 2 + 2 = 5. 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 100001;
long long n, m;
long long arr[MAXN];
long long sum[4 * MAXN];
long long Max[4 * MAXN];
void build(int node = 1, int l = 0, int r = n - 1) {
  if (l == r) {
    sum[node] = arr[l];
    Max[node] = arr[l];
    return;
  }
  int m = (l + r) / 2;
  build(2 * node, l, m);
  build(2 * node + 1, m + 1, r);
  sum[node] = sum[2 * node] + sum[2 * node + 1];
  Max[node] = max(Max[2 * node], Max[2 * node + 1]);
}
void update(int idx, long long val, int node = 1, int l = 0, int r = n - 1) {
  if (r < idx || idx < l) return;
  if (l == r) {
    arr[l] = val;
    sum[node] = val;
    Max[node] = val;
    return;
  } else {
    int m = (l + r) / 2;
    update(idx, val, 2 * node, l, m);
    update(idx, val, 2 * node + 1, m + 1, r);
  }
  sum[node] = sum[2 * node] + sum[2 * node + 1];
  Max[node] = max(Max[2 * node], Max[2 * node + 1]);
}
void update2(int a, int b, long long mod, int node = 1, int l = 0,
             int r = n - 1) {
  int m = (l + r) / 2;
  if (a <= l && r <= b) {
    if (Max[node] >= mod && l == r) {
      Max[node] = Max[node] % mod;
      sum[node] = Max[node];
      arr[l] = Max[node];
    } else if (Max[node] >= mod) {
      update2(a, b, mod, 2 * node, l, m);
      update2(a, b, mod, 2 * node + 1, m + 1, r);
      sum[node] = sum[2 * node] + sum[2 * node + 1];
      Max[node] = max(Max[2 * node], Max[2 * node + 1]);
    }
    return;
  }
  if (a <= m) {
    update2(a, b, mod, 2 * node, l, m);
  }
  if (b > m) {
    update2(a, b, mod, 2 * node + 1, m + 1, r);
  }
  sum[node] = sum[2 * node] + sum[2 * node + 1];
  Max[node] = max(Max[2 * node], Max[2 * node + 1]);
}
long long query_sum(int a, int b, int node = 1, int l = 0, int r = n - 1) {
  if (a > r || b < l) return 0;
  if (a <= l && r <= b) return sum[node];
  int m = (l + r) / 2;
  long long ret = 0;
  long long q1 = query_sum(a, b, 2 * node, l, m);
  long long q2 = query_sum(a, b, 2 * node + 1, m + 1, r);
  ret = q1 + q2;
  return ret;
}
int main() {
  cin >> n >> m;
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
  }
  build();
  for (int i = 0; i < m; i++) {
    int arch;
    cin >> arch;
    if (arch == 1) {
      int l, r;
      cin >> l >> r;
      cout << query_sum(l - 1, r - 1) << '\n';
    } else if (arch == 2) {
      int l, r;
      long long x;
      cin >> l >> r >> x;
      update2(l - 1, r - 1, x);
    } else if (arch == 3) {
      int k;
      long long x;
      cin >> k >> x;
      update(k - 1, x);
    }
  }
  return 0;
}