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 mx = 3e5 + 5;
int Set(int N, int pos) { return N = N | (1 << pos); }
int reset(int N, int pos) { return N = N & ~(1 << pos); }
bool check(int N, int pos) { return (bool)(N & (1 << pos)); }
struct data {
  long long sum;
  long long maxi;
} tree[mx << 2];
int arr[mx];
void build(int node, int b, int e) {
  if (b == e) {
    tree[node].sum = tree[node].maxi = arr[b];
    return;
  }
  int mid = (b + e) >> 1;
  int left = node << 1;
  int right = left | 1;
  build(left, b, mid);
  build(right, mid + 1, e);
  tree[node].sum = tree[left].sum + tree[right].sum;
  tree[node].maxi = max(tree[left].maxi, tree[right].maxi);
}
void update1(int node, int b, int e, int pos, int val) {
  if (pos > e || pos < b) return;
  if (pos == e && b == e) {
    tree[node].maxi = tree[node].sum = val;
    return;
  }
  int mid = (b + e) >> 1;
  int left = node << 1;
  int right = left | 1;
  update1(left, b, mid, pos, val);
  update1(right, mid + 1, e, pos, val);
  tree[node].sum = tree[left].sum + tree[right].sum;
  tree[node].maxi = max(tree[left].maxi, tree[right].maxi);
}
void update2(int node, int b, int e, int l, int r, int mod) {
  if (l > e || r < b) return;
  if (tree[node].maxi < mod) return;
  if (b == e && b >= l && e <= r) {
    tree[node].sum = tree[node].sum % mod;
    tree[node].maxi = tree[node].maxi % mod;
    return;
  }
  int mid = (b + e) >> 1;
  int left = node << 1;
  int right = left | 1;
  update2(left, b, mid, l, r, mod);
  update2(right, mid + 1, e, l, r, mod);
  tree[node].sum = tree[left].sum + tree[right].sum;
  tree[node].maxi = max(tree[left].maxi, tree[right].maxi);
}
long long query(int node, int b, int e, int l, int r) {
  if (l > e || r < b) return 0;
  if (b >= l && e <= r) return tree[node].sum;
  int mid = (b + e) >> 1;
  int left = node << 1;
  int right = left | 1;
  long long x = query(left, b, mid, l, r);
  long long y = query(right, mid + 1, e, l, r);
  return x + y;
}
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d", &arr[i]);
  build(1, 1, n);
  while (m--) {
    int tp;
    scanf("%d", &tp);
    if (tp == 1) {
      int l, r;
      scanf("%d%d", &l, &r);
      long long ans = query(1, 1, n, l, r);
      printf("%lld\n", ans);
    }
    if (tp == 2) {
      int l, r, x;
      scanf("%d%d%d", &l, &r, &x);
      update2(1, 1, n, l, r, x);
    }
    if (tp == 3) {
      int pos, val;
      scanf("%d%d", &pos, &val);
      update1(1, 1, n, pos, val);
    }
  }
  return 0;
}