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;
const int N = 101000;
long long calc(long long x) { return x * (x + 1) / 2; }
struct Sgt {
  long long sum[N << 2], ll[N << 2], rr[N << 2];
  void up(int l, int r, int rt) {
    int mid = l + r >> 1;
    sum[rt] = sum[rt << 1] + sum[rt << 1 | 1] - calc(rr[rt << 1]) -
              calc(ll[rt << 1 | 1]) + calc(rr[rt << 1] + ll[rt << 1 | 1]);
    ll[rt] = ll[rt << 1] + (ll[rt << 1] == mid - l + 1 ? ll[rt << 1 | 1] : 0);
    rr[rt] = rr[rt << 1 | 1] + (rr[rt << 1 | 1] == r - mid ? rr[rt << 1] : 0);
  }
  void modify(int p, int v, int l, int r, int rt) {
    if (l == r) {
      sum[rt] = ll[rt] = rr[rt] = v;
      return;
    }
    int mid = l + r >> 1;
    if (p <= mid)
      modify(p, v, l, mid, rt << 1);
    else
      modify(p, v, mid + 1, r, rt << 1 | 1);
    up(l, r, rt);
  }
} tree[20];
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 0; i < n; i++) {
    int c;
    scanf("%d", &c);
    for (int j = 0; j < 20; j++)
      if (c >> j & 1) tree[j].modify(i, 1, 0, n - 1, 1);
  }
  while (m--) {
    int a, b;
    scanf("%d%d", &a, &b);
    a--;
    long long ans = 0;
    for (int j = 0; j < 20; j++) {
      tree[j].modify(a, b >> j & 1, 0, n - 1, 1);
      ans += tree[j].sum[1] << j;
    }
    printf("%lld\n", ans);
  }
  return 0;
}