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 N = 2e5;
const int oo = 1e9 + 5;
int du[] = {-1, 0, 0, 1};
int dv[] = {0, -1, 1, 0};
const long long mod = 1e9 + 7;
long long sqr(long long x) { return x * x; }
int n, q;
struct Node {
  long long sum, Max;
  int vt;
} f[4 * N + 5];
long long lz[4 * N + 5], a[N + 5];
void build(int id, int l, int r) {
  if (l > r) return;
  if (l == r) {
    f[id].sum = a[l];
    f[id].Max = a[l];
    f[id].vt = l;
    return;
  }
  int mid = (l + r) / 2;
  build(id * 2, l, mid);
  build(id * 2 + 1, mid + 1, r);
  f[id].sum = f[id * 2].sum + f[id * 2 + 1].sum;
  if (f[id * 2].Max > f[id * 2 + 1].Max) {
    f[id].Max = f[id * 2].Max;
    f[id].vt = f[id * 2].vt;
  } else {
    f[id].Max = f[id * 2 + 1].Max;
    f[id].vt = f[id * 2 + 1].vt;
  }
}
void update1(int id, int l, int r, int u, int v, long long val) {
  if (l > r || l > v || r < u) return;
  if (l == r) {
    a[l] %= val;
    f[id].sum = a[l];
    f[id].Max = a[l];
    f[id].vt = l;
    return;
  }
  int mid = (l + r) / 2;
  if (f[id * 2].Max >= val) update1(id * 2, l, mid, u, v, val);
  if (f[id * 2 + 1].Max >= val) update1(id * 2 + 1, mid + 1, r, u, v, val);
  f[id].sum = f[id * 2].sum + f[id * 2 + 1].sum;
  if (f[id * 2].Max > f[id * 2 + 1].Max) {
    f[id].Max = f[id * 2].Max;
    f[id].vt = f[id * 2].vt;
  } else {
    f[id].Max = f[id * 2 + 1].Max;
    f[id].vt = f[id * 2 + 1].vt;
  }
}
void update2(int id, int l, int r, int u) {
  if (l > r || l > u || r < u) return;
  if (l == r) {
    f[id].sum = a[l];
    f[id].Max = a[l];
    f[id].vt = l;
    return;
  }
  int mid = (l + r) / 2;
  update2(id * 2, l, mid, u);
  update2(id * 2 + 1, mid + 1, r, u);
  f[id].sum = f[id * 2].sum + f[id * 2 + 1].sum;
  if (f[id * 2].Max > f[id * 2 + 1].Max) {
    f[id].Max = f[id * 2].Max;
    f[id].vt = f[id * 2].vt;
  } else {
    f[id].Max = f[id * 2 + 1].Max;
    f[id].vt = f[id * 2 + 1].vt;
  }
}
long long getsum(int id, int l, int r, int u, int v) {
  if (l > r || l > v || r < u) return 0;
  if (u <= l && r <= v) return f[id].sum;
  int mid = (l + r) / 2;
  return getsum(id * 2, l, mid, u, v) + getsum(id * 2 + 1, mid + 1, r, u, v);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> q;
  for (int i = 1; i <= n; i++) cin >> a[i];
  build(1, 1, n);
  for (int i = 1; i <= q; i++) {
    int c;
    cin >> c;
    if (c == 1) {
      int u, v;
      cin >> u >> v;
      cout << getsum(1, 1, n, u, v) << endl;
    } else if (c == 2) {
      int u, v;
      long long val;
      cin >> u >> v >> val;
      update1(1, 1, n, u, v, val);
    } else {
      int u;
      long long x;
      cin >> u >> x;
      a[u] = x;
      update2(1, 1, n, u);
    }
  }
  return 0;
}