/* TASK: Train Journey (trains)
* Task proposal by: Bohdan Feysa (UKR)
* Task prepared by: Bohdan Feysa (UKR), Zigmas Bitinas (LTU SC), Aldas Lenkšas (LTU SC)
*
* Task can be solved by square root decomposition. Let's choose a constant C = sqrt(N).
* Now we do the Dynamic Programming solution, where dp[i] represents how many journies
* can be made if we would end the journey at the i-th city.
*
* If d[i] >= C, then each dp[j] is increased by dp[i], where j = i + t * d[i]. This takes
* O(N/C) time.
*
* If d[i] < C, then we store the value in the array, so that we can add this value later.
*
*/
#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9+7;
const int MAXN = 1e5;
const int C = 400; // ~sqrt(MAXN)
int n;
long long d[MAXN], x[MAXN];
int dp[MAXN];
int acc[C][C]; // the value to add to dp
int addToAcc[MAXN][C]; // the value to add to accumulator
void addToAccumulator(long long i, int jump, int a) {
// a is the value to add
if (i >= n)
return; // not interested
if (a < 0) // deal with subtractions
a = (a + MOD) % MOD;
addToAcc[i][jump] = (addToAcc[i][jump] + a) % MOD;
}
int main() {
// read input
cin >> n;
for (int i = 0; i < n; i++)
cin >> d[i] >> x[i];
dp[0] = 1; // base case
for (int i = 0; i < n; i++) {
for (int jump = 1; jump < C; jump++) {
int m = i % jump;
acc[jump][m] = (acc[jump][m] + addToAcc[i][jump]) % MOD;
dp[i] = (dp[i] + acc[jump][m]) % MOD;
}
if (d[i] == 0) continue; // not working teleporter
// sqrt decomposition
if (d[i] < C) {
int jump = d[i];
addToAccumulator(i + jump, jump, dp[i]);
addToAccumulator(i + jump * x[i] + jump, jump, -dp[i]);
} else {
for (int j = i + d[i], xu = 1; xu <= x[i] && j < n; xu++, j += d[i]) {
dp[j] = (dp[j] + dp[i]) % MOD;
}
}
}
// ans = dp[0] + dp[1] + ... + dp[n-1]
int ans = 0;
for (int i = 0; i < n; i++)
ans = (ans + dp[i]) % MOD;
cout << (ans % MOD) << endl;
return 0;
}
