#include <bits/stdc++.h>
using namespace std;
using ll = long long;
struct SegTree {
int n;
struct Node { ll mn, lazy; };
vector<Node> st;
SegTree(int _n): n(_n), st(4*n+4, {0,0}) {}
void apply(int p, ll v) {
st[p].mn += v;
st[p].lazy += v;
}
void push(int p) {
if (st[p].lazy) {
apply(p<<1, st[p].lazy);
apply(p<<1|1, st[p].lazy);
st[p].lazy = 0;
}
}
void pull(int p) {
st[p].mn = min(st[p<<1].mn, st[p<<1|1].mn);
}
// range add v to [L,R]
void update(int p, int l, int r, int L, int R, ll v) {
if (L > r || R < l) return;
if (L <= l && r <= R) {
apply(p, v);
return;
}
push(p);
int m = (l + r) >> 1;
update(p<<1, l, m, L, R, v);
update(p<<1|1, m+1, r, L, R, v);
pull(p);
}
void update(int L, int R, ll v) { update(1, 0, n, L, R, v); }
// point query at idx
ll query(int p, int l, int r, int idx) {
if (l == r) return st[p].mn;
push(p);
int m = (l + r) >> 1;
return idx <= m
? query(p<<1, l, m, idx)
: query(p<<1|1, m+1, r, idx);
}
ll query(int idx) { return query(1, 0, n, idx); }
// find first position >= x in [l..r] where value < val
int first_below(int p, int l, int r, int x, ll val) {
if (r < x || st[p].mn >= val) return -1;
if (l == r) return l;
push(p);
int m = (l + r) >> 1;
int res = first_below(p<<1, l, m, x, val);
if (res != -1) return res;
return first_below(p<<1|1, m+1, r, x, val);
}
int first_below(int x, ll val) { return first_below(1, 0, n, x, val); }
};
int main(){
ios::sync_with_stdio(false);
cin.tie(nullptr);
int n;
cin >> n;
vector<int> h(n), k(n);
for(int i = 0; i < n; i++)
cin >> h[i] >> k[i];
// process in increasing height order
vector<int> ord(n);
iota(ord.begin(), ord.end(), 0);
sort(ord.begin(), ord.end(), [&](int a, int b){
return h[a] < h[b];
});
int H = *max_element(h.begin(), h.end());
SegTree st(H);
// exactly the same logic as your BIT version:
// let val = st.query(last)
// find idx1 = first position < val on [0..h[i]]
// find idx2 = first position < val+1 on [0..h[i]]
// do two range-adds to simulate adding 1 to [idx1..h[i]] but NOT overshooting
for(int i: ord) {
int last = h[i] - k[i];
ll val = st.query(last); // how many sails already at 'last'
int idx1 = st.first_below(0, val+1); // first pos < val+1 => first < (val+1) == ≤ val
int idx2 = st.first_below(0, val+2); // first pos < val+2 => first < (val+2) == ≤ val+1
// now do two range adds:
// 1) +1 on [idx1 .. h[i]]
// 2) -1 on [idx2 .. h[i]]
st.update(idx1, h[i], +1);
st.update(idx2, h[i], -1);
}
// collect answer
ll res = 0;
for(int i = 0; i <= H; i++){
ll c = st.query(i);
res += c * (c-1) / 2;
}
cout << res << "\n";
return 0;
}
