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 S[17][100000];
int v[100000];
inline void update(int bit, int p, int v) {
  while (p <= 100000) {
    S[bit][p] += v;
    p += p & -p;
  }
}
long long Sum(int bit, int p) {
  long long sum = 0;
  while (p > 0) {
    sum += S[bit][p];
    p -= p & -p;
  }
  return sum;
}
inline long long calc(long long n) { return ((n * (n + 1)) >> 1); }
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &v[i]);
    for (int j = 0; j < 17; ++j) update(j, i + 1, (v[i] >> j) & 1);
  }
  long long sum = 0;
  for (int i = 0; i < 17; ++i) {
    for (int j = 0; j < n; ++j) {
      int len;
      for (len = 0; j + len < n; ++len)
        if (((v[j + len] >> i) & 1) == 0) break;
      sum += calc(len) * (1 << i);
      j += len;
    }
  }
  for (int i = 0; i < m; ++i) {
    int p, val;
    scanf("%d%d", &p, &val);
    --p;
    int crr = v[p];
    int next = val;
    v[p] = next;
    for (int j = 0; j < 17; ++j, crr >>= 1, next >>= 1) {
      if ((crr & 1) == (next & 1)) continue;
      if (next & 1) update(j, p + 1, 1);
      long long s1 = Sum(j, p + 1);
      int l1, r1, m1;
      pair<int, int> p1;
      l1 = -1;
      r1 = p;
      while (r1 - l1 > 1) {
        m1 = (l1 + r1) / 2;
        if (s1 - Sum(j, m1) == (p - m1) + 1)
          r1 = m1;
        else
          l1 = m1;
      }
      l1 = r1;
      r1 = p;
      int l2, r2, m2;
      pair<int, int> p2;
      l2 = p;
      r2 = n;
      while (r2 - l2 > 1) {
        m2 = (l2 + r2) / 2;
        if (Sum(j, m2 + 1) - s1 == (m2 - p))
          l2 = m2;
        else
          r2 = m2;
      }
      r2 = l2;
      l2 = p;
      if ((next & 1) == 0) update(j, p + 1, -1);
      if (next & 1) {
        sum += (calc(r2 - l1 + 1) - calc(r1 - l1) - calc(r2 - l2)) * (1 << j);
      } else {
        sum += (-calc(r2 - l1 + 1) + calc(r1 - l1) + calc(r2 - l2)) * (1 << j);
      }
    }
    printf("%I64d\n", sum);
  }
  return 0;
}