#include<bits/stdc++.h>

using namespace std;

struct SegNode
{
  int v;

  SegNode(int v) : v(v) {}

  SegNode operator*(const SegNode &r) const
  {
    return (v < r.v ? *this : r);
  }
} e(1 << 30);

struct SegmentTree
{
  int sz;
  vector< SegNode > seg;

  SegmentTree(int n)
  {
    sz = 1;
    while(sz < n) sz <<= 1;
    seg.assign(2 * sz - 1, e);
  }

  void update(int k, const SegNode &x)
  {
    k += sz - 1;
    seg[k] = x;
    while(k > 0) {
      k = (k - 1) >> 1;
      seg[k] = seg[2 * k + 1] * seg[2 * k + 2];
    }
  }

  SegNode query(int a, int b, int k, int l, int r)
  {
    if(a >= r || b <= l) return (e);
    if(a <= l && r <= b) return (seg[k]);
    return (query(a, b, 2 * k + 1, l, (l + r) >> 1) * query(a, b, 2 * k + 2, (l + r) >> 1, r));
  }

  SegNode query(int a, int b)
  {
    return (query(a, b, 0, 0, sz));
  }
};

int main()
{
  int N, Q, B[200000];
  vector< int > ev[200000];

  scanf("%d", &N);
  int add = 0;
  for(int i = 0; i < N; i++) {
    scanf("%d", &B[i]);
    if(B[i] == 0) ++add, B[i] = -1;
  }
  scanf("%d", &Q);
  for(int i = 0; i < Q; i++) {
    int l, r;
    scanf("%d %d", &l, &r);
    --l;
    ev[l].emplace_back(r);
  }

  SegmentTree tree(N);
  int dp[200001]; // 区間 [0, i) を考えた時の最小コスト
  fill_n(dp, 200001, 1 << 30);
  dp[0] = 0;
  for(int i = 0; i < N; i++) {
    for(auto &j : ev[i]) {
      int v = min(dp[j], min(dp[i], tree.query(i, j).v));
      tree.update(j, v);
      dp[j] = v;
    }
    dp[i + 1] = min(dp[i + 1], dp[i] + B[i]);
  }
  printf("%d\n", dp[N] + add);
}

### Prompt

Construct a CPP code solution to the problem outlined:

You are given a sequence a = \\{a_1, ..., a_N\\} with all zeros, and a sequence b = \\{b_1, ..., b_N\\} consisting of 0 and 1. The length of both is N.

You can perform Q kinds of operations. The i-th operation is as follows:

* Replace each of a_{l_i}, a_{l_i + 1}, ..., a_{r_i} with 1.



Minimize the hamming distance between a and b, that is, the number of i such that a_i \neq b_i, by performing some of the Q operations.

Constraints

* 1 \leq N \leq 200,000
* b consists of 0 and 1.
* 1 \leq Q \leq 200,000
* 1 \leq l_i \leq r_i \leq N
* If i \neq j, either l_i \neq l_j or r_i \neq r_j.

Input

Input is given from Standard Input in the following format:


N
b_1 b_2 ... b_N
Q
l_1 r_1
l_2 r_2
:
l_Q r_Q


Output

Print the minimum possible hamming distance.

Examples

Input

3
1 0 1
1
1 3


Output

1


Input

3
1 0 1
2
1 1
3 3


Output

0


Input

3
1 0 1
2
1 1
2 3


Output

1


Input

5
0 1 0 1 0
1
1 5


Output

2


Input

9
0 1 0 1 1 1 0 1 0
3
1 4
5 8
6 7


Output

3


Input

15
1 1 0 0 0 0 0 0 1 0 1 1 1 0 0
9
4 10
13 14
1 7
4 14
9 11
2 6
7 8
3 12
7 13


Output

5


Input

10
0 0 0 1 0 0 1 1 1 0
7
1 4
2 5
1 3
6 7
9 9
1 5
7 9


Output

1

### Response

```cpp
#include<bits/stdc++.h>

using namespace std;

struct SegNode
{
  int v;

  SegNode(int v) : v(v) {}

  SegNode operator*(const SegNode &r) const
  {
    return (v < r.v ? *this : r);
  }
} e(1 << 30);

struct SegmentTree
{
  int sz;
  vector< SegNode > seg;

  SegmentTree(int n)
  {
    sz = 1;
    while(sz < n) sz <<= 1;
    seg.assign(2 * sz - 1, e);
  }

  void update(int k, const SegNode &x)
  {
    k += sz - 1;
    seg[k] = x;
    while(k > 0) {
      k = (k - 1) >> 1;
      seg[k] = seg[2 * k + 1] * seg[2 * k + 2];
    }
  }

  SegNode query(int a, int b, int k, int l, int r)
  {
    if(a >= r || b <= l) return (e);
    if(a <= l && r <= b) return (seg[k]);
    return (query(a, b, 2 * k + 1, l, (l + r) >> 1) * query(a, b, 2 * k + 2, (l + r) >> 1, r));
  }

  SegNode query(int a, int b)
  {
    return (query(a, b, 0, 0, sz));
  }
};

int main()
{
  int N, Q, B[200000];
  vector< int > ev[200000];

  scanf("%d", &N);
  int add = 0;
  for(int i = 0; i < N; i++) {
    scanf("%d", &B[i]);
    if(B[i] == 0) ++add, B[i] = -1;
  }
  scanf("%d", &Q);
  for(int i = 0; i < Q; i++) {
    int l, r;
    scanf("%d %d", &l, &r);
    --l;
    ev[l].emplace_back(r);
  }

  SegmentTree tree(N);
  int dp[200001]; // 区間 [0, i) を考えた時の最小コスト
  fill_n(dp, 200001, 1 << 30);
  dp[0] = 0;
  for(int i = 0; i < N; i++) {
    for(auto &j : ev[i]) {
      int v = min(dp[j], min(dp[i], tree.query(i, j).v));
      tree.update(j, v);
      dp[j] = v;
    }
    dp[i + 1] = min(dp[i + 1], dp[i] + B[i]);
  }
  printf("%d\n", dp[N] + add);
}

```