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;
pair<long long, long long> t[5000000];
pair<long long, long long> lz[5000000];
int n, nb, n1, m, k, l, r;
int so;
bool test;
string s;
set<int> st;
map<int, int> mapp;
void merged(int node, int l = 0, int r = 0) {
  t[node].first = t[2 * node].first + t[2 * node + 1].first;
  t[node].second = max(t[2 * node].second, t[2 * node + 1].second);
}
void go(int node, int l, int r, int d) {
  if (t[node].second < d) return;
  if (l == r) {
    t[node].first %= d;
    t[node].second %= d;
    return;
  }
  int mid = (l + r) / 2;
  go(2 * node, l, mid, d);
  go(2 * node + 1, mid + 1, r, d);
  merged(node, l, r);
}
void update1(int node, int l1, int r1, int l, int r, int d) {
  if ((l1 > r) || (r1 < l)) return;
  if ((l <= l1) && (r1 <= r)) {
    go(node, l1, r1, d);
    return;
  }
  int mid = (l1 + r1) / 2;
  update1(2 * node, l1, mid, l, r, d);
  update1(2 * node + 1, mid + 1, r1, l, r, d);
  merged(node, l1, r1);
}
void update2(int a, int d) {
  t[nb + a] = make_pair(d, d);
  int k = (nb + a) / 2;
  while (k) {
    merged(k);
    k >>= 1;
  }
}
long long query(int node, int l1, int r1, int l, int r) {
  if ((l1 > r) || (r1 < l)) return 0ll;
  if ((l <= l1) && (r1 <= r)) {
    return t[node].first;
  }
  int mid = (l1 + r1) / 2;
  long long a = query(2 * node, l1, mid, l, r);
  long long b = query(2 * node + 1, mid + 1, r1, l, r);
  merged(node, l1, r1);
  return a + b;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n >> m;
  nb = 1;
  int k = n - 1;
  while (k) {
    k >>= 1, nb <<= 1;
  }
  for (int i = 0; i < n; i++) {
    int a;
    cin >> a;
    t[nb + i] = make_pair(a, a);
  }
  for (int i = nb - 1; i > 0; i--) merged(i);
  for (int i = 0; i < m; i++) {
    int a, b, c, d;
    cin >> a;
    if (a == 1) {
      cin >> b >> c;
      cout << query(1, 0, nb - 1, b - 1, c - 1) << '\n';
    } else if (a == 2) {
      cin >> b >> c >> d;
      update1(1, 0, nb - 1, b - 1, c - 1, d);
    } else {
      cin >> b >> c;
      update2(b - 1, c);
    }
  }
}