Problem: Inna is fed up with jokes about female logic. So she started using binary logic instead.

Inna has an array of n elements a1[1], a1[2], ..., a1[n]. Girl likes to train in her binary logic, so she does an exercise consisting of n stages: on the first stage Inna writes out all numbers from array a1, on the i-th (i ≥ 2) stage girl writes all elements of array ai, which consists of n - i + 1 integers; the k-th integer of array ai is defined as follows: ai[k] = ai - 1[k] AND ai - 1[k + 1]. Here AND is bit-wise binary logical operation.

Dima decided to check Inna's skill. He asks Inna to change array, perform the exercise and say the sum of all <image> elements she wrote out during the current exercise.

Help Inna to answer the questions!

Input

The first line contains two integers n and m (1 ≤ n, m ≤ 105) — size of array a1 and number of Dima's questions. Next line contains n integers a1[1], a1[2], ..., a1[n] (0 ≤ ai ≤ 105) — initial array elements.

Each of next m lines contains two integers — Dima's question description. Each question consists of two integers pi, vi (1 ≤ pi ≤ n; 0 ≤ vi ≤ 105). For this question Inna should make a1[pi] equals vi, and then perform the exercise. Please, note that changes are saved from question to question.

Output

For each question print Inna's answer on a single line.

Examples

Input

3 4
1 1 1
1 1
2 2
3 2
1 2


Output

6
4
7
12

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int a[100005];
struct no {
  long long pre, suf, sum, len;
  no(long long pre, long long suf, long long sum)
      : pre(pre), suf(suf), sum(sum) {}
  no() {}
};
no tree[400005][17];
no merge(no a, no b) {
  no c = no(0, 0, 0);
  c.suf = b.suf == b.len ? a.suf + b.suf : b.suf;
  c.pre = a.pre == a.len ? a.pre + b.pre : a.pre;
  c.len = a.len + b.len;
  c.sum = a.sum + b.sum + a.suf * b.pre;
  return c;
}
void update(int bit, int node, int l, int r, int pos, int val) {
  if (l > pos || r < pos || l > r) return;
  if (l == r && pos == l) {
    tree[node][bit].pre = val;
    tree[node][bit].sum = val;
    tree[node][bit].suf = val;
    tree[node][bit].len = 1;
    return;
  }
  int mid = (l + r) >> 1;
  update(bit, 2 * node, l, mid, pos, val);
  update(bit, 2 * node + 1, mid + 1, r, pos, val);
  tree[node][bit] = merge(tree[node * 2][bit], tree[2 * node + 1][bit]);
}
int main() {
  int n, m;
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    for (int j = 0; j < 17; j++) {
      bool val = a[i] & (1 << j);
      update(j, 1, 1, n, i, val);
    }
  }
  while (m--) {
    int p, b;
    scanf("%d %d", &p, &b);
    long long ans = 0;
    for (int j = 0; j < 17; j++) {
      bool val = b & (1 << j);
      update(j, 1, 1, n, p, val);
      ans += tree[1][j].sum * (1 << j);
    }
    printf("%lld\n", ans);
    a[p] = b;
  }
  return 0;
}