#include <bits/stdc++.h>
using namespace std;
int n, m, q, l, r, ma, mi;
int timer;
vector<vector<long long> > g, u, num;
vector<long long> used, tin, fup;
vector<long long> pre, sum, z;
long long res, len;
void dfs(int v, int p = -1) {
  used[v] = 1;
  tin[v] = fup[v] = timer++;
  for (int i = 0; i < g[v].size(); ++i) {
    int to = g[v][i];
    if (to == p) continue;
    if (used[to])
      fup[v] = min(fup[v], tin[to]);
    else {
      dfs(to, v);
      fup[v] = min(fup[v], fup[to]);
      if (fup[to] > tin[v]) {
        u[v][i] = 0;
        u[to][num[v][i]] = 0;
      }
    }
  }
}
void dfss(int v) {
  used[v] = 1;
  mi = min(mi, v);
  ma = max(ma, v);
  for (int i = 0; i < g[v].size(); ++i) {
    if (!u[v][i]) continue;
    int to = g[v][i];
    if (!used[to]) dfss(to);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> m;
  g.resize(n);
  u.resize(n);
  num.resize(n);
  for (int i = 0; i < m; ++i) {
    int a, b;
    cin >> a >> b;
    --a;
    --b;
    num[a].push_back(u[b].size());
    num[b].push_back(u[a].size());
    g[a].push_back(b);
    g[b].push_back(a);
    u[a].push_back(1);
    u[b].push_back(1);
  }
  timer = 0;
  used.assign(n, 0);
  tin.resize(n);
  fup.resize(n);
  for (int i = 0; i < n; ++i) {
    if (used[i]) continue;
    dfs(i);
  }
  z.assign(n, -1);
  used.assign(n, 0);
  for (int i = 0; i < n; ++i) {
    if (used[i]) continue;
    mi = n + 1;
    ma = -1;
    dfss(i);
    if (mi != ma) z[ma] = max(z[ma], (long long)mi);
  }
  pre.assign(n, 0);
  for (int i = 1; i < n; ++i) {
    pre[i] = z[i] + 1;
    pre[i] = max(pre[i], pre[i - 1]);
  }
  sum.assign(n, 1);
  for (int i = 1; i < n; ++i) sum[i] = sum[i - 1] + (i - pre[i] + 1);
  cin >> q;
  while (q--) {
    cin >> l >> r;
    --l;
    --r;
    len = (r - l + 1);
    if (pre[r] <= l)
      res = len * (len + 1) / 2;
    else if (pre[l] >= l)
      res = sum[r] - (l ? sum[l - 1] : 0);
    else {
      int le = l;
      int ri = r;
      while (ri - le > 1) {
        int mid = (le + ri) / 2;
        (pre[mid] < l ? le : ri) = mid;
      }
      len = (le - l + 1);
      res = len * (len + 1) / 2 + sum[r] - sum[le];
    }
    cout << res << "\n";
  }
  return 0;
}

### Prompt

In Cpp, your task is to solve the following problem:

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;
int n, m, q, l, r, ma, mi;
int timer;
vector<vector<long long> > g, u, num;
vector<long long> used, tin, fup;
vector<long long> pre, sum, z;
long long res, len;
void dfs(int v, int p = -1) {
  used[v] = 1;
  tin[v] = fup[v] = timer++;
  for (int i = 0; i < g[v].size(); ++i) {
    int to = g[v][i];
    if (to == p) continue;
    if (used[to])
      fup[v] = min(fup[v], tin[to]);
    else {
      dfs(to, v);
      fup[v] = min(fup[v], fup[to]);
      if (fup[to] > tin[v]) {
        u[v][i] = 0;
        u[to][num[v][i]] = 0;
      }
    }
  }
}
void dfss(int v) {
  used[v] = 1;
  mi = min(mi, v);
  ma = max(ma, v);
  for (int i = 0; i < g[v].size(); ++i) {
    if (!u[v][i]) continue;
    int to = g[v][i];
    if (!used[to]) dfss(to);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> m;
  g.resize(n);
  u.resize(n);
  num.resize(n);
  for (int i = 0; i < m; ++i) {
    int a, b;
    cin >> a >> b;
    --a;
    --b;
    num[a].push_back(u[b].size());
    num[b].push_back(u[a].size());
    g[a].push_back(b);
    g[b].push_back(a);
    u[a].push_back(1);
    u[b].push_back(1);
  }
  timer = 0;
  used.assign(n, 0);
  tin.resize(n);
  fup.resize(n);
  for (int i = 0; i < n; ++i) {
    if (used[i]) continue;
    dfs(i);
  }
  z.assign(n, -1);
  used.assign(n, 0);
  for (int i = 0; i < n; ++i) {
    if (used[i]) continue;
    mi = n + 1;
    ma = -1;
    dfss(i);
    if (mi != ma) z[ma] = max(z[ma], (long long)mi);
  }
  pre.assign(n, 0);
  for (int i = 1; i < n; ++i) {
    pre[i] = z[i] + 1;
    pre[i] = max(pre[i], pre[i - 1]);
  }
  sum.assign(n, 1);
  for (int i = 1; i < n; ++i) sum[i] = sum[i - 1] + (i - pre[i] + 1);
  cin >> q;
  while (q--) {
    cin >> l >> r;
    --l;
    --r;
    len = (r - l + 1);
    if (pre[r] <= l)
      res = len * (len + 1) / 2;
    else if (pre[l] >= l)
      res = sum[r] - (l ? sum[l - 1] : 0);
    else {
      int le = l;
      int ri = r;
      while (ri - le > 1) {
        int mid = (le + ri) / 2;
        (pre[mid] < l ? le : ri) = mid;
      }
      len = (le - l + 1);
      res = len * (len + 1) / 2 + sum[r] - sum[le];
    }
    cout << res << "\n";
  }
  return 0;
}

```