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 = 2 * 1e5 + 10;
int N, M;
long long arr[MX];
pair<long long, long long> st[4 * MX];
void merge(int L, int R, int p) {
  st[p].first = st[p * 2].first + st[p * 2 + 1].first;
  int lc = st[p * 2].second, rc = st[p * 2 + 1].second;
  if (arr[lc] >= arr[rc])
    st[p].second = lc;
  else
    st[p].second = rc;
}
void B(int L, int R, int p) {
  if (L == R) {
    st[p].first = arr[L];
    st[p].second = L;
    return;
  }
  int mid = (L + R) / 2;
  B(L, mid, p * 2);
  B(mid + 1, R, p * 2 + 1);
  merge(L, R, p);
}
pair<long long, long long> Q(int L, int R, int p, int i, int j) {
  if (L > R || i > R || j < L) return {0, -1};
  if (L >= i && R <= j) {
    return st[p];
  }
  pair<long long, long long> q1, q2;
  int mid = L + R;
  mid /= 2;
  q1 = Q(L, mid, p * 2, i, j);
  q2 = Q(mid + 1, R, p * 2 + 1, i, j);
  if (q2.second == -1)
    return {q1.first + q2.first, q1.second};
  else if (q1.second == -1)
    return {q1.first + q2.first, q2.second};
  else if (arr[q2.second] >= arr[q1.second])
    return {q1.first + q2.first, q2.second};
  return {q1.first + q2.first, q1.second};
}
void U(int L, int R, int p, int i, int j, int val) {
  if (L > R || i > R || j < L) return;
  if (L >= i && R <= j) {
    st[p].first = val;
    if (L == R) arr[L] = val;
    return;
  }
  int mid = L + R;
  mid /= 2;
  U(L, mid, p * 2, i, j, val);
  U(mid + 1, R, p * 2 + 1, i, j, val);
  merge(L, R, p);
}
void up(int L, int R, int x) {
  pair<long long, long long> ans = Q(0, N - 1, 1, L, R);
  if (arr[ans.second] < x) return;
  U(0, N - 1, 1, ans.second, ans.second, arr[ans.second] % x);
  up(L, R, x);
}
int main() {
  scanf("%d%d", &N, &M);
  for (int i = 0; i < N; i++) {
    int x;
    scanf("%d", &x);
    arr[i] = (long long)x;
  }
  B(0, N - 1, 1);
  for (int i = 0; i < M; i++) {
    int t;
    scanf("%d", &t);
    if (t == 1) {
      int l, r;
      scanf("%d%d", &l, &r);
      cout << Q(0, N - 1, 1, l - 1, r - 1).first << "\n";
    } else if (t == 2) {
      int l, r, x;
      scanf("%d%d%d", &l, &r, &x);
      up(l - 1, r - 1, x);
    } else {
      int k, x;
      scanf("%d%d", &k, &x);
      U(0, N - 1, 1, k - 1, k - 1, x);
    }
  }
}