#include <bits/stdc++.h>
using namespace std;
const int N = 3e5 + 7;
int n, m;
vector<int> g[N];
void eadd(int u, int v) {
  g[u].push_back(v);
  g[v].push_back(u);
}
int t = 0;
long long int mx[N];
long long int psum[N];
int dt[N], lw[N];
stack<int> stk;
void dfs(int u, int p = -1) {
  dt[u] = lw[u] = ++t;
  stk.push(u);
  for (int v : g[u])
    if (!dt[v]) {
      dfs(v, u);
      lw[u] = min(lw[u], lw[v]);
    } else if (v != p)
      lw[u] = min(lw[u], dt[v]);
  if (lw[u] == dt[u]) {
    if (u == stk.top()) {
      stk.pop();
      return;
    }
    int cmn = u, cmx = u, i = u;
    do {
      i = stk.top();
      stk.pop();
      cmn = min(cmn, i);
      cmx = max(cmx, i);
    } while (i != u);
    mx[cmn] = cmx - 1;
  }
}
void print(string s, long long int val) {
  cout << '[' << s << ": " << val << ']';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  cin >> n >> m;
  for (int i = 1; i <= m; ++i) {
    int u, v;
    cin >> u >> v;
    eadd(u, v);
  }
  fill(mx, mx + n + 1, n);
  for (int i = 1; i <= n; ++i)
    if (!dt[i]) dfs(i);
  for (int i = n - 1; i; --i) mx[i] = min(mx[i], mx[i + 1]);
  for (int i = 1; i <= n; ++i) psum[i] = psum[i - 1] + mx[i];
  int q;
  cin >> q;
  while (q--) {
    long long int l, r;
    cin >> l >> r;
    long long int t1 = (r * (r + 1ll) - (l - 1) * l) >> 1, t2 = r - l + 1ll,
                  t01 = 0, t02 = 0;
    long long int k = upper_bound(mx + l, mx + r + 1, r) - mx;
    --k;
    if (k >= l) t01 = (psum[k] - psum[l - 1]);
    ++k;
    if (k <= r) t02 = (r - k + 1ll) * r;
    cout << t01 + t02 - t1 + t2 << '\n';
  }
  return 0;
}

### Prompt

Develop a solution in CPP 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 N = 3e5 + 7;
int n, m;
vector<int> g[N];
void eadd(int u, int v) {
  g[u].push_back(v);
  g[v].push_back(u);
}
int t = 0;
long long int mx[N];
long long int psum[N];
int dt[N], lw[N];
stack<int> stk;
void dfs(int u, int p = -1) {
  dt[u] = lw[u] = ++t;
  stk.push(u);
  for (int v : g[u])
    if (!dt[v]) {
      dfs(v, u);
      lw[u] = min(lw[u], lw[v]);
    } else if (v != p)
      lw[u] = min(lw[u], dt[v]);
  if (lw[u] == dt[u]) {
    if (u == stk.top()) {
      stk.pop();
      return;
    }
    int cmn = u, cmx = u, i = u;
    do {
      i = stk.top();
      stk.pop();
      cmn = min(cmn, i);
      cmx = max(cmx, i);
    } while (i != u);
    mx[cmn] = cmx - 1;
  }
}
void print(string s, long long int val) {
  cout << '[' << s << ": " << val << ']';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  cin >> n >> m;
  for (int i = 1; i <= m; ++i) {
    int u, v;
    cin >> u >> v;
    eadd(u, v);
  }
  fill(mx, mx + n + 1, n);
  for (int i = 1; i <= n; ++i)
    if (!dt[i]) dfs(i);
  for (int i = n - 1; i; --i) mx[i] = min(mx[i], mx[i + 1]);
  for (int i = 1; i <= n; ++i) psum[i] = psum[i - 1] + mx[i];
  int q;
  cin >> q;
  while (q--) {
    long long int l, r;
    cin >> l >> r;
    long long int t1 = (r * (r + 1ll) - (l - 1) * l) >> 1, t2 = r - l + 1ll,
                  t01 = 0, t02 = 0;
    long long int k = upper_bound(mx + l, mx + r + 1, r) - mx;
    --k;
    if (k >= l) t01 = (psum[k] - psum[l - 1]);
    ++k;
    if (k <= r) t02 = (r - k + 1ll) * r;
    cout << t01 + t02 - t1 + t2 << '\n';
  }
  return 0;
}

```