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 maxn = 100000 + 10;
int num[maxn];
struct SegTree {
  int sum[maxn << 2], pre[maxn << 2], suff[maxn << 2];
  long long val[maxn << 2];
  int bitp;
  void PushUp(int l, int r, int rt) {
    int m = (l + r) >> 1;
    if (pre[rt << 1] == m - l + 1)
      pre[rt] = m - l + 1 + pre[rt << 1 | 1];
    else
      pre[rt] = pre[rt << 1];
    if (suff[rt << 1 | 1] == r - m)
      suff[rt] = suff[rt << 1] + r - m;
    else
      suff[rt] = suff[rt << 1 | 1];
    if (suff[rt << 1] && pre[rt << 1 | 1]) {
      val[rt] = val[rt << 1] + val[rt << 1 | 1];
      long long tmp = suff[rt << 1];
      val[rt] -= tmp * (tmp + 1) / 2;
      tmp = pre[rt << 1 | 1];
      val[rt] -= tmp * (tmp + 1) / 2;
      tmp = suff[rt << 1] + pre[rt << 1 | 1];
      val[rt] += tmp * (tmp + 1) / 2;
    } else
      val[rt] = val[rt << 1] + val[rt << 1 | 1];
  }
  void build(int l, int r, int rt) {
    if (l == r) {
      val[rt] = sum[rt] = pre[rt] = suff[rt] = (num[l] >> bitp) & 1;
      return;
    }
    int m = (l + r) >> 1;
    build(l, m, rt << 1);
    build(m + 1, r, rt << 1 | 1);
    PushUp(l, r, rt);
  }
  void Update(int p, int l, int r, int rt, int v) {
    if (l == r) {
      sum[rt] = val[rt] = pre[rt] = suff[rt] = v;
      return;
    }
    int m = (l + r) >> 1;
    if (m >= p)
      Update(p, l, m, rt << 1, v);
    else
      Update(p, m + 1, r, rt << 1 | 1, v);
    PushUp(l, r, rt);
  }
} tree[17];
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; ++i) scanf("%d", &num[i]);
  for (int i = 0; i < 17; ++i) {
    tree[i].bitp = i;
    tree[i].build(1, n, 1);
  }
  int p, v;
  long long sum;
  while (m--) {
    scanf("%d%d", &p, &v);
    sum = 0;
    for (int i = 0; i < 17; ++i) {
      tree[i].Update(p, 1, n, 1, (v >> i) & 1);
      sum += (1 << i) * tree[i].val[1];
    }
    printf("%I64d\n", sum);
  }
  return 0;
}