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

template <class T>
struct SegTree {
    using F = function<T(T, T)>; // operator type
    const int n; // array size
    const T id; // identity element
    F op; //operator
    vector<T> dat; // data array
    SegTree(int n, T id, F op) :n(n), id(id), op(op), dat(n * 2, id) {}
    T operator[](int i) { return dat[i + n]; }
    void modify(int i, const T& v) { // set v at position i
        dat[i += n] = v;
        while(i >>= 1) dat[i] = op(dat[i * 2], dat[i * 2 + 1]);
    }
    T query(int l, int r) { // query on interval [l, r)
        T a{ id }, b{ id };
        for(l += n, r += n; l < r; l >>= 1, r >>= 1) {
            if(l & 1) a = op(a, dat[l++]);
            if(r & 1) b = op(dat[r - 1], b);
        }
        return op(a, b);
    }
};

constexpr int N_MAX{ 200002 };

SegTree<int> st(N_MAX, numeric_limits<int>::max(), [](int a, int b) { return min(a, b); });
int N, Q, b[N_MAX], c[N_MAX], ans;
set<int> q[N_MAX];
int main() {
    cin.sync_with_stdio(0); cin.tie(0);
    cin >> N;
    for(int i{ 1 }; i <= N; ++i) {
        cin >> b[i];
        c[i] = c[i - 1] + !b[i];
    }
    cin >> Q;
    for(int i{}; i < Q; ++i) {
        int l, r; cin >> l >> r;
        q[l].insert(r);
    }
    for(int i{ 1 }; i <= N; ++i) {
        for(auto& x : q[i])
            st.modify(x, min(st.query(i, x + 1), ans + c[N] - c[i - 1]));
        ans = min(ans + b[i], st[i] - c[N] + c[i]);
    }
    cout << ans << endl;
}

### Prompt

Please formulate 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;

template <class T>
struct SegTree {
    using F = function<T(T, T)>; // operator type
    const int n; // array size
    const T id; // identity element
    F op; //operator
    vector<T> dat; // data array
    SegTree(int n, T id, F op) :n(n), id(id), op(op), dat(n * 2, id) {}
    T operator[](int i) { return dat[i + n]; }
    void modify(int i, const T& v) { // set v at position i
        dat[i += n] = v;
        while(i >>= 1) dat[i] = op(dat[i * 2], dat[i * 2 + 1]);
    }
    T query(int l, int r) { // query on interval [l, r)
        T a{ id }, b{ id };
        for(l += n, r += n; l < r; l >>= 1, r >>= 1) {
            if(l & 1) a = op(a, dat[l++]);
            if(r & 1) b = op(dat[r - 1], b);
        }
        return op(a, b);
    }
};

constexpr int N_MAX{ 200002 };

SegTree<int> st(N_MAX, numeric_limits<int>::max(), [](int a, int b) { return min(a, b); });
int N, Q, b[N_MAX], c[N_MAX], ans;
set<int> q[N_MAX];
int main() {
    cin.sync_with_stdio(0); cin.tie(0);
    cin >> N;
    for(int i{ 1 }; i <= N; ++i) {
        cin >> b[i];
        c[i] = c[i - 1] + !b[i];
    }
    cin >> Q;
    for(int i{}; i < Q; ++i) {
        int l, r; cin >> l >> r;
        q[l].insert(r);
    }
    for(int i{ 1 }; i <= N; ++i) {
        for(auto& x : q[i])
            st.modify(x, min(st.query(i, x + 1), ans + c[N] - c[i - 1]));
        ans = min(ans + b[i], st[i] - c[N] + c[i]);
    }
    cout << ans << endl;
}

```