#include <bits/stdc++.h>
using namespace std;

int N, Q;

// 区間を左端でソートして順にセグメント木を更新

template <typename T>
class SegTree
{
    int N;
    int head;
    vector<T> x;
    T defaultT;
    using func = function<T(const T&, const T&)>;
    func tree_update;

    T _query_search(int begin, int end, int pos, int l, int r)
    {
        if (r <= begin || l >= end) return defaultT;
        if (l >= begin && r <= end) return x[pos];
        T v_l = _query_search(begin, end, 2*pos+1, l, (l+r)/2);
        T v_r = _query_search(begin, end, 2*pos+2, (l+r)/2, r);
        return tree_update(v_l, v_r);
    }

public:
    SegTree(int N, T val_default, func f) : N(N), defaultT(val_default), tree_update(f)
    {
        int N_tmp = 1;
        while (N_tmp < N) N_tmp <<= 1;
        x = vector<T>(N_tmp*2, defaultT);
        head = N_tmp - 1;
    }

    void update(int pos, T val)
    {
        pos += head;
        x[pos] = val;
        while (pos)
        {
            pos = (pos - 1) / 2;
            x[pos] = tree_update(x[pos*2+1], x[pos*2+2]);
        }
    }
    T query(int begin, int end)
    {
        return _query_search(begin, end, 0, 0, (int)x.size() / 2);
    }
};

int main()
{
    cin >> N;
    vector<int> b(N);
    for (auto &v : b) cin >> v;
    SegTree<int> RMQ(N+1, 200000, [](int a, int b) { return min(a, b); });
    RMQ.update(0, 0);

    vector<vector<int>> c(N+1);
    cin >> Q;
    for (int i = 0; i < Q; i++)
    {
        int tmp1, tmp2; cin >> tmp1 >> tmp2;
        c[tmp1].push_back(tmp2);
    }

    for (int l = 1; l <= N; l++)
    {
        for (auto r : c[l]) RMQ.update(r, RMQ.query(l - 1, r + 1));
        RMQ.update(l, min(RMQ.query(l-1, l) + (b[l-1] ? 1 : -1), RMQ.query(l, l+1)));
    }
    cout << RMQ.query(N, N + 1) + N - accumulate(b.begin(), b.end(), 0) << endl;
}

### Prompt

Your challenge is to write 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 <bits/stdc++.h>
using namespace std;

int N, Q;

// 区間を左端でソートして順にセグメント木を更新

template <typename T>
class SegTree
{
    int N;
    int head;
    vector<T> x;
    T defaultT;
    using func = function<T(const T&, const T&)>;
    func tree_update;

    T _query_search(int begin, int end, int pos, int l, int r)
    {
        if (r <= begin || l >= end) return defaultT;
        if (l >= begin && r <= end) return x[pos];
        T v_l = _query_search(begin, end, 2*pos+1, l, (l+r)/2);
        T v_r = _query_search(begin, end, 2*pos+2, (l+r)/2, r);
        return tree_update(v_l, v_r);
    }

public:
    SegTree(int N, T val_default, func f) : N(N), defaultT(val_default), tree_update(f)
    {
        int N_tmp = 1;
        while (N_tmp < N) N_tmp <<= 1;
        x = vector<T>(N_tmp*2, defaultT);
        head = N_tmp - 1;
    }

    void update(int pos, T val)
    {
        pos += head;
        x[pos] = val;
        while (pos)
        {
            pos = (pos - 1) / 2;
            x[pos] = tree_update(x[pos*2+1], x[pos*2+2]);
        }
    }
    T query(int begin, int end)
    {
        return _query_search(begin, end, 0, 0, (int)x.size() / 2);
    }
};

int main()
{
    cin >> N;
    vector<int> b(N);
    for (auto &v : b) cin >> v;
    SegTree<int> RMQ(N+1, 200000, [](int a, int b) { return min(a, b); });
    RMQ.update(0, 0);

    vector<vector<int>> c(N+1);
    cin >> Q;
    for (int i = 0; i < Q; i++)
    {
        int tmp1, tmp2; cin >> tmp1 >> tmp2;
        c[tmp1].push_back(tmp2);
    }

    for (int l = 1; l <= N; l++)
    {
        for (auto r : c[l]) RMQ.update(r, RMQ.query(l - 1, r + 1));
        RMQ.update(l, min(RMQ.query(l-1, l) + (b[l-1] ? 1 : -1), RMQ.query(l, l+1)));
    }
    cout << RMQ.query(N, N + 1) + N - accumulate(b.begin(), b.end(), 0) << endl;
}

```