/*
PROBLEM: Count Valid Subarrays Without Forbidden Pairs Inside
 
Given:
- N positions (1 to N)
- M pairs (u, v)
 
Task: Count how many subarrays [l, r] don't contain any pair FULLY INSIDE.
 
A pair (u, v) is "fully inside" a subarray [l, r] if:
- BOTH u and v are in the range [l, r]
- Example: pair (2, 4) is fully inside [1, 5] because both 2 and 4 are in [1, 5]
 
A subarray is VALID if no pair is fully contained inside it. This means:
- For every pair (a, b), at least one endpoint must be outside [l, r]
 
Example:
N=5, M=2
Pairs: (1, 3) and (2, 5)
 
Invalid subarrays: [1,3], [1,4], [1,5], [2,5] (pair fully inside)
Valid subarrays: [1,1], [2,2], [3,3], [4,4], [5,5], [1,2], [2,3], [3,4], [4,5], [2,4], [3,5]
 
Key Insight:
For a subarray [l, r] to be valid:
- For every pair (a, b) where b <= r, we must have a < l (left endpoint outside)
- In other words: l must be > the maximum left endpoint of all pairs ending at or before r
 
Approach:
1. For each right endpoint r, track the maximum left endpoint among all pairs ending at r
2. As we extend r, maintain the leftmost valid starting position (bound)
3. Count valid subarrays ending at r: all positions from bound+1 to r can be left endpoints
 
Time: O(N + M)
*/
 
#include <bits/stdc++.h>
using namespace std;
 
int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
 
    int n, m;
    cin >> n >> m;
 
    // max[r] = maximum left endpoint among all pairs with right endpoint r
    vector<int> mx(n + 1, 0);
 
    for (int i = 0; i < m; i++) {
        int u, v;
        cin >> u >> v;
 
        // Make sure l <= r
        int l = min(u, v);
        int r = max(u, v);
 
        // Track the furthest left endpoint for each right endpoint
        mx[r] = max(mx[r], l);
    }
 
    long long ans = 0;
 
    // bound = leftmost position where pair would be fully inside
    int bound = 0;
 
    // Try each position as the right endpoint
    for (int r = 1; r <= n; r++) {
        // Update bound: if any pair ends at r with left >= bound, update it
        bound = max(bound, mx[r]);
 
        // Count subarrays ending at r
        // We can start from any position > bound up to r
        // That's (r - bound) choices
        ans += (long long)(r - bound);
    }
 
    cout << ans << "\n";
 
    return 0;
}

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