#include <bits/stdc++.h>
using namespace std;
const int M = 3e5 + 5;
vector<int> vec[M];
int color[M];
long long segTree[4 * M], lazy[4 * M];
int lef[M], safe_left[M];
stack<int> stck;
vector<pair<int, int> > query[M];
long long answer[M];
void dfs(int u, int p) {
  color[u] = 1;
  stck.push(u);
  for (int i = 0; i < vec[u].size(); i++) {
    int v = vec[u][i];
    if (v == p || color[v] == 2) continue;
    if (color[v] == 0) {
      dfs(v, u);
    } else {
      int lo = M, hi = 0;
      while (stck.size()) {
        int x = stck.top();
        stck.pop();
        lo = min(lo, x);
        hi = max(hi, x);
        if (x == v) break;
      }
      lef[hi] = lo;
    }
  }
  if (stck.size() && stck.top() == u) stck.pop();
  color[u] = 2;
}
void push_down(int node, int l, int r) {
  segTree[node] += (r - l + 1) * lazy[node];
  if (l != r) {
    lazy[2 * node] += lazy[node];
    lazy[2 * node + 1] += lazy[node];
  }
  lazy[node] = 0;
}
void update(int node, int l, int r, int ql, int qr) {
  push_down(node, l, r);
  if (l > qr || l > r || r < ql) return;
  if (l >= ql && r <= qr) {
    lazy[node]++;
    push_down(node, l, r);
    return;
  }
  int mid = (l + r) / 2;
  update(2 * node, l, mid, ql, qr);
  update(2 * node + 1, mid + 1, r, ql, qr);
  segTree[node] = segTree[2 * node] + segTree[2 * node + 1];
}
long long ask(int node, int l, int r, int ql, int qr) {
  push_down(node, l, r);
  if (l > qr || l > r || r < ql) return 0;
  if (l >= ql && r <= qr) {
    return segTree[node];
  }
  int mid = (l + r) / 2;
  long long A = ask(2 * node, l, mid, ql, qr);
  long long B = ask(2 * node + 1, mid + 1, r, ql, qr);
  return (A + B);
}
int main() {
  int N, M;
  scanf("%d", &N);
  scanf("%d", &M);
  for (int i = 1; i <= M; i++) {
    int u, v;
    scanf("%d", &u);
    scanf("%d", &v);
    vec[u].push_back(v);
    vec[v].push_back(u);
  }
  for (int i = 1; i <= N; i++) {
    if (color[i]) continue;
    dfs(i, 0);
  }
  int till = 0;
  for (int i = 1; i <= N; i++) {
    till = max(till, lef[i]);
    safe_left[i] = till + 1;
  }
  int Q;
  scanf("%d", &Q);
  for (int i = 1; i <= Q; i++) {
    int L, R;
    scanf("%d", &L);
    scanf("%d", &R);
    query[R].push_back(make_pair(L, i));
  }
  for (int i = 1; i <= N; i++) {
    int lef_limit = safe_left[i];
    update(1, 1, N, lef_limit, i);
    for (int j = 0; j < query[i].size(); j++) {
      int LL = query[i][j].first;
      int RR = i;
      answer[query[i][j].second] = ask(1, 1, N, LL, RR);
    }
  }
  for (int i = 1; i <= Q; i++) {
    printf("%lld\n", answer[i]);
  }
}

### Prompt

Please provide a cpp coded solution to the problem described below:

You are given an undirected graph with n vertices. There are no edge-simple cycles with the even length in it. In other words, there are no cycles of even length that pass each edge at most once. Let's enumerate vertices from 1 to n. 

You have to answer q queries. Each query is described by a segment of vertices [l; r], and you have to count the number of its subsegments [x; y] (l ≤ x ≤ y ≤ r), such that if we delete all vertices except the segment of vertices [x; y] (including x and y) and edges between them, the resulting graph is bipartite.

Input

The first line contains two integers n and m (1 ≤ n ≤ 3·105, 1 ≤ m ≤ 3·105) — the number of vertices and the number of edges in the graph.

The next m lines describe edges in the graph. The i-th of these lines contains two integers ai and bi (1 ≤ ai, bi ≤ n; ai ≠ bi), denoting an edge between vertices ai and bi. It is guaranteed that this graph does not contain edge-simple cycles of even length.

The next line contains a single integer q (1 ≤ q ≤ 3·105) — the number of queries.

The next q lines contain queries. The i-th of these lines contains two integers li and ri (1 ≤ li ≤ ri ≤ n) — the query parameters.

Output

Print q numbers, each in new line: the i-th of them should be the number of subsegments [x; y] (li ≤ x ≤ y ≤ ri), such that the graph that only includes vertices from segment [x; y] and edges between them is bipartite.

Examples

Input

6 6
1 2
2 3
3 1
4 5
5 6
6 4
3
1 3
4 6
1 6


Output

5
5
14


Input

8 9
1 2
2 3
3 1
4 5
5 6
6 7
7 8
8 4
7 2
3
1 8
1 4
3 8


Output

27
8
19

Note

The first example is shown on the picture below:

<image>

For the first query, all subsegments of [1; 3], except this segment itself, are suitable.

For the first query, all subsegments of [4; 6], except this segment itself, are suitable.

For the third query, all subsegments of [1; 6] are suitable, except [1; 3], [1; 4], [1; 5], [1; 6], [2; 6], [3; 6], [4; 6].

The second example is shown on the picture below:

<image>

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int M = 3e5 + 5;
vector<int> vec[M];
int color[M];
long long segTree[4 * M], lazy[4 * M];
int lef[M], safe_left[M];
stack<int> stck;
vector<pair<int, int> > query[M];
long long answer[M];
void dfs(int u, int p) {
  color[u] = 1;
  stck.push(u);
  for (int i = 0; i < vec[u].size(); i++) {
    int v = vec[u][i];
    if (v == p || color[v] == 2) continue;
    if (color[v] == 0) {
      dfs(v, u);
    } else {
      int lo = M, hi = 0;
      while (stck.size()) {
        int x = stck.top();
        stck.pop();
        lo = min(lo, x);
        hi = max(hi, x);
        if (x == v) break;
      }
      lef[hi] = lo;
    }
  }
  if (stck.size() && stck.top() == u) stck.pop();
  color[u] = 2;
}
void push_down(int node, int l, int r) {
  segTree[node] += (r - l + 1) * lazy[node];
  if (l != r) {
    lazy[2 * node] += lazy[node];
    lazy[2 * node + 1] += lazy[node];
  }
  lazy[node] = 0;
}
void update(int node, int l, int r, int ql, int qr) {
  push_down(node, l, r);
  if (l > qr || l > r || r < ql) return;
  if (l >= ql && r <= qr) {
    lazy[node]++;
    push_down(node, l, r);
    return;
  }
  int mid = (l + r) / 2;
  update(2 * node, l, mid, ql, qr);
  update(2 * node + 1, mid + 1, r, ql, qr);
  segTree[node] = segTree[2 * node] + segTree[2 * node + 1];
}
long long ask(int node, int l, int r, int ql, int qr) {
  push_down(node, l, r);
  if (l > qr || l > r || r < ql) return 0;
  if (l >= ql && r <= qr) {
    return segTree[node];
  }
  int mid = (l + r) / 2;
  long long A = ask(2 * node, l, mid, ql, qr);
  long long B = ask(2 * node + 1, mid + 1, r, ql, qr);
  return (A + B);
}
int main() {
  int N, M;
  scanf("%d", &N);
  scanf("%d", &M);
  for (int i = 1; i <= M; i++) {
    int u, v;
    scanf("%d", &u);
    scanf("%d", &v);
    vec[u].push_back(v);
    vec[v].push_back(u);
  }
  for (int i = 1; i <= N; i++) {
    if (color[i]) continue;
    dfs(i, 0);
  }
  int till = 0;
  for (int i = 1; i <= N; i++) {
    till = max(till, lef[i]);
    safe_left[i] = till + 1;
  }
  int Q;
  scanf("%d", &Q);
  for (int i = 1; i <= Q; i++) {
    int L, R;
    scanf("%d", &L);
    scanf("%d", &R);
    query[R].push_back(make_pair(L, i));
  }
  for (int i = 1; i <= N; i++) {
    int lef_limit = safe_left[i];
    update(1, 1, N, lef_limit, i);
    for (int j = 0; j < query[i].size(); j++) {
      int LL = query[i][j].first;
      int RR = i;
      answer[query[i][j].second] = ask(1, 1, N, LL, RR);
    }
  }
  for (int i = 1; i <= Q; i++) {
    printf("%lld\n", answer[i]);
  }
}

```