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;
int mo[4][2] = {-1, 0, 1, 0, 0, -1, 0, 1};
const int maxn = 1e5 + 10;
long long n, m, is = 0, c = 0, ans = 0, p[maxn];
struct xx {
  long long l, r, w, maxnn, add;
} t[maxn << 2];
void up(long long k) {
  t[k].w = t[k << 1].w + t[k << 1 | 1].w;
  t[k].maxnn = max(t[k << 1].maxnn, t[k << 1 | 1].maxnn);
}
void down(long long k) {
  if (t[k].add) {
    t[k << 1].add += t[k].add;
    t[k << 1 | 1].add += t[k].add;
    t[k << 1].w += t[k].add * (t[k << 1].r - t[k << 1].l + 1);
    t[k << 1 | 1].w += t[k].add * (t[k << 1 | 1].r - t[k << 1 | 1].l + 1);
    t[k].add = 0;
  }
}
void build(long long k, long long l, long long r) {
  t[k].l = l, t[k].r = r, t[k].w = t[k].maxnn = 0;
  if (l == r) {
    t[k].w = t[k].maxnn = p[l];
    return;
  }
  long long mid = (l + r) >> 1;
  build(k << 1, l, mid);
  build(k << 1 | 1, mid + 1, r);
  up(k);
}
void update(long long k, long long c, long long v) {
  if (t[k].l == t[k].r) {
    t[k].w = t[k].maxnn = v;
    return;
  }
  down(k);
  long long mid = (t[k].l + t[k].r) >> 1;
  if (c <= mid)
    update(k << 1, c, v);
  else
    update(k << 1 | 1, c, v);
  up(k);
}
long long getsum(long long k, long long l, long long r) {
  if (l <= t[k].l && t[k].r <= r) return t[k].w;
  down(k);
  long long mid = (t[k].l + t[k].r) >> 1;
  long long sum = 0;
  if (l <= mid) sum += getsum(k << 1, l, r);
  if (r > mid) sum += getsum(k << 1 | 1, l, r);
  return sum;
}
void updatemod(long long k, long long l, long long r, long long mod) {
  if (t[k].maxnn < mod) return;
  if (t[k].l == t[k].r) {
    t[k].w %= mod;
    t[k].maxnn %= mod;
    return;
  }
  down(k);
  long long mid = (t[k].l + t[k].r) >> 1;
  if (l <= mid) updatemod(k << 1, l, r, mod);
  if (r > mid) updatemod(k << 1 | 1, l, r, mod);
  up(k);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  cout << fixed << setprecision(2);
  scanf("%lld%lld", &n, &m);
  for (long long i = 1; i <= n; i++) scanf("%lld", &p[i]);
  build(1, 1, n);
  while (m--) {
    long long op;
    scanf("%lld", &op);
    if (op == 1) {
      long long a, b;
      scanf("%lld%lld", &a, &b);
      printf("%lld\n", getsum(1, a, b));
    } else if (op == 2) {
      long long a, b, c;
      scanf("%lld%lld%lld", &a, &b, &c);
      updatemod(1, a, b, c);
    } else if (op == 3) {
      long long a, b;
      scanf("%lld%lld", &a, &b);
      update(1, a, b);
    }
  }
  return 0;
}