#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;

class SegmentTree {
  public:
    SegmentTree(int size, int value):
        m_size(size),
        m_data(size * 2, value) {}

    void update(int position, int value) {
        for (position += m_size; position; position /= 2)
            m_data[position] = min(m_data[position], value);
    }

    int query(int from, int to) {
        int answer = numeric_limits<int>::max();
        for (from += m_size, to += m_size; from < to; from /= 2, to /= 2) {
            if (from & 1)
                answer = min(answer, m_data[from++]);
            if (to & 1)
                answer = min(answer, m_data[--to]);
        }
        return answer;
    }

  private:
    int m_size;
    vector<int> m_data;
};

int main() {
    ios::sync_with_stdio(false);
    int N; cin >> N;
    vector<int> B(N + 1, 0);
    for (int i = 1; i <= N; ++i)
        cin >> B[i];

    vector<int> partial(N + 1, 0);
    for (int i = 1; i <= N; ++i)
        partial[i] = partial[i - 1] + B[i];

    int Q; cin >> Q;
    vector< vector<int> > rights(N + 1);
    for (int i = 0; i < Q; ++i) {
        int x, y; cin >> x >> y;
        rights[x].emplace_back(y);
    }

    SegmentTree S(N + 1, N);
    vector<int> dp(N + 1, N);
    dp[0] = 0;
    for (int i = 1; i <= N; ++i) {
        dp[i] = min(dp[i], dp[i - 1] + B[i]);
        S.update(i - 1, dp[i - 1] - (i - 1) + partial[i - 1]);
        // dp[i] = min(dp[j] + (i - j) - (partial[i] - partial[j]))
        //       = i - partial[i] + min(dp[j] - j + partial[j])
        for (auto &right : rights[i]) {
            int best = right - partial[right] + S.query(i - 1, right);
            dp[right] = min(dp[right], best);
            S.update(right, best - right + partial[right]);
        }
    }

    cout << dp[N] << "\n";
}

### Prompt

Your task is to create a cpp solution to the following problem:

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 <iostream>
#include <vector>
#include <algorithm>

using namespace std;

class SegmentTree {
  public:
    SegmentTree(int size, int value):
        m_size(size),
        m_data(size * 2, value) {}

    void update(int position, int value) {
        for (position += m_size; position; position /= 2)
            m_data[position] = min(m_data[position], value);
    }

    int query(int from, int to) {
        int answer = numeric_limits<int>::max();
        for (from += m_size, to += m_size; from < to; from /= 2, to /= 2) {
            if (from & 1)
                answer = min(answer, m_data[from++]);
            if (to & 1)
                answer = min(answer, m_data[--to]);
        }
        return answer;
    }

  private:
    int m_size;
    vector<int> m_data;
};

int main() {
    ios::sync_with_stdio(false);
    int N; cin >> N;
    vector<int> B(N + 1, 0);
    for (int i = 1; i <= N; ++i)
        cin >> B[i];

    vector<int> partial(N + 1, 0);
    for (int i = 1; i <= N; ++i)
        partial[i] = partial[i - 1] + B[i];

    int Q; cin >> Q;
    vector< vector<int> > rights(N + 1);
    for (int i = 0; i < Q; ++i) {
        int x, y; cin >> x >> y;
        rights[x].emplace_back(y);
    }

    SegmentTree S(N + 1, N);
    vector<int> dp(N + 1, N);
    dp[0] = 0;
    for (int i = 1; i <= N; ++i) {
        dp[i] = min(dp[i], dp[i - 1] + B[i]);
        S.update(i - 1, dp[i - 1] - (i - 1) + partial[i - 1]);
        // dp[i] = min(dp[j] + (i - j) - (partial[i] - partial[j]))
        //       = i - partial[i] + min(dp[j] - j + partial[j])
        for (auto &right : rights[i]) {
            int best = right - partial[right] + S.query(i - 1, right);
            dp[right] = min(dp[right], best);
            S.update(right, best - right + partial[right]);
        }
    }

    cout << dp[N] << "\n";
}

```