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 n, m, x;
long long Summa;
set<int> Zeros[18], Ones[18];
set<int>::iterator IterateOverElements;
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 0; i < 18; i++) {
    Zeros[i].insert(n), Ones[i].insert(n);
    Zeros[i].insert(-1), Ones[i].insert(-1);
  }
  for (int i = 0; i < n; i++) {
    scanf("%d", &x);
    int k = 1;
    int index = -1;
    while (true) {
      index++;
      if (index == 18) break;
      if (k & x)
        Ones[index].insert(i);
      else
        Zeros[index].insert(i);
      k <<= 1;
    }
  }
  long long Val = 1;
  for (int i = 0; i < 18; i++) {
    int posBegin = 0;
    while (true) {
      posBegin = *Ones[i].lower_bound(posBegin);
      if (posBegin == n) {
        break;
      }
      IterateOverElements = Zeros[i].upper_bound(posBegin);
      long long Right = *IterateOverElements;
      long long Left = (*--IterateOverElements);
      Summa += Val * (Right - Left - 1) * (Right - Left) / 2;
      posBegin = *Zeros[i].lower_bound(posBegin);
    }
    Val *= 2;
  }
  for (int i = 0; i < m; i++) {
    int position, v;
    scanf("%d %d", &position, &v);
    --position;
    long long k = 1;
    int index = -1;
    while (1) {
      index++;
      if (index == 18) break;
      if (k & v) {
        IterateOverElements = Ones[index].lower_bound(position);
        if (*IterateOverElements != position) {
          set<int>::iterator m_it = Zeros[index].lower_bound(position);
          IterateOverElements = m_it;
          long long Right = *++m_it;
          long long Middle = position;
          long long Left = *--IterateOverElements;
          Summa += k * (Right - Left - 1) * (Right - Left) / 2 -
                   k * (Middle - Left - 1) * (Middle - Left) / 2 -
                   k * (Right - Middle - 1) * (Right - Middle) / 2;
          Zeros[index].erase(position);
          Ones[index].insert(position);
        }
      } else {
        IterateOverElements = Zeros[index].lower_bound(position);
        if (*IterateOverElements != position) {
          long long Middle = position;
          long long Right = *IterateOverElements;
          long long Left = *--IterateOverElements;
          Summa -= k * (Right - Left - 1) * (Right - Left) / 2 -
                   k * (Right - Middle - 1) * (Right - Middle) / 2 -
                   k * (Middle - Left - 1) * (Middle - Left) / 2;
          Ones[index].erase(position);
          Zeros[index].insert(position);
        }
      }
      k <<= 1;
    }
    printf("%I64d\n", Summa);
  }
  return 0;
}