// Slower C++ solution:
#include<bits/stdc++.h>

#define rep(i,a,b) for(int i=a;i<b;i++)

using namespace std; void _main(); int main() { cin.tie(0); ios::sync_with_stdio(false); _main(); }

//---------------------------------------------------------------------------------------------------

int mod = 1000000007;

int add(int x, int y) { return (x += y) >= mod ? x - mod : x; }

template<class... T> int add(int x, T... y) { return add(x, add(y...)); }

int mul(int x, int y) { return 1LL * x * y % mod; }

template<class... T> int mul(int x, T... y) { return mul(x, mul(y...)); }

int sub(int x, int y) { return add(x, mod - y); }

int modpow(int a, long long b) {

    int ret = 1; while (b > 0) {

        if (b & 1) ret = 1LL * ret * a % mod; a = 1LL * a * a % mod; b >>= 1;

    } return ret;

}

int modinv(int a) { return modpow(a, mod - 2); }

#define def 0

template<class V, int NV> struct SegTree {

    V comp(V l, V r) { return add(l,r); };

    vector<V> val; SegTree() { val = vector<V>(NV * 2, def); }

    V get(int l, int r) { //[l,r]

        if (l > r) return def;

        l += NV; r += NV + 1; V ret = def;

        while (l < r) { if (l & 1) ret = comp(ret, val[l++]); if (r & 1) ret = comp(ret, val[--r]); l /= 2; r /= 2; }

        return ret;

    }

    V get(int x) { return val[x + NV]; }

    void update(int i, V v) { i += NV; val[i] = v; while (i>1) i >>= 1, val[i] = comp(val[i * 2], val[i * 2 + 1]); }

};

/*---------------------------------------------------------------------------------------------------

　　　　　　　　　　　 ∧＿∧  

　　　　　 ∧＿∧ 　（´<_｀ ）　 Welcome to My Coding Space!

　　　　 （ ´_ゝ`）　/　 ⌒i     

　　　　／　　　＼　 　  |　|     

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　 ＿_(__ﾆつ/　    ＿/ .| .|＿＿＿＿  

　 　　　＼/＿＿＿＿/　（u　⊃  

---------------------------------------------------------------------------------------------------*/









int N;

pair<int, int> P[201010];

//---------------------------------------------------------------------------------------------------

void zaatsu() {

    vector<int> dic;

    rep(i, 0, N) dic.push_back(P[i].second);

    sort(dic.begin(), dic.end());

    dic.erase(unique(dic.begin(), dic.end()), dic.end());



    rep(i, 0, N) P[i].second = lower_bound(dic.begin(), dic.end(), P[i].second) - dic.begin() + 1;



    /*rep(i, 0, N) printf("%d %d\n", L[i], R[i]);

    printf("<%d>\n", M);*/

}

//---------------------------------------------------------------------------------------------------

#define INF INT_MAX/2

#define rrep(i,a,b) for(int i=a;i>=b;i--)

int L[201010], R[201010];

void change() {

    sort(P, P + N);



    rep(i, 0, N) {

        int pre = 0;

        if (0 < i) pre = R[i - 1];

        R[i] = max(pre, P[i].second);

    }

    

    L[N] = INF;

    rrep(i, N - 1, 0) {

        int pre = L[i + 1];

        L[i] = min(pre, P[i].second);

    }



    //rep(i, 0, N) printf("%d %d\n", L[i], R[i]);

}

//---------------------------------------------------------------------------------------------------

// Solution1

/*int dp[2020][2020];

int dodp() {

    dp[0][0] = 1;

    rep(i, 0, N) {

        rep(j, 0, M + 1) dp[i + 1][j] = dp[i][j];



        int sm = 0;

        rep(j, L[i] - 1, R[i] + 1) sm = add(sm, dp[i][j]);

        dp[i + 1][R[i]] = add(dp[i + 1][R[i]], sm);

    }



    return dp[N][N];

}*/

//---------------------------------------------------------------------------------------------------

// Solution2

/*int dp[201010];

int dodp() {

    dp[0] = 1;

    rep(i, 0, N) {

        int sm = 0;

        rep(j, L[i] - 1, R[i] + 1) sm = add(sm, dp[j]);

        dp[R[i]] = add(dp[R[i]], sm);

    }



    return dp[N];

}*/

//---------------------------------------------------------------------------------------------------

SegTree<int, 1 << 18> dp;

int dodp() {

    dp.update(0, 1);

    rep(i, 0, N) {

        int sm = dp.get(L[i] - 1, R[i]);

        dp.update(R[i], add(dp.get(R[i]), sm));

    }



    return dp.get(N);

}

//---------------------------------------------------------------------------------------------------

void _main() {

    cin >> N;

    rep(i, 0, N) cin >> P[i].first >> P[i].second;



    zaatsu();

    change();



    cout << dodp() << endl;

}

// Faster C++ solution for the same problem (measured speedup: 3.082631337670488x):
#include<bits/stdc++.h>

#define rep(i,a,b) for(int i=a;i<b;i++)

using namespace std; void _main(); int main() { cin.tie(0); ios::sync_with_stdio(false); _main(); }

//---------------------------------------------------------------------------------------------------

int mod = 1000000007;

int add(int x, int y) { return (x += y) >= mod ? x - mod : x; }

template<class... T> int add(int x, T... y) { return add(x, add(y...)); }

int mul(int x, int y) { return 1LL * x * y % mod; }

template<class... T> int mul(int x, T... y) { return mul(x, mul(y...)); }

int sub(int x, int y) { return add(x, mod - y); }

int modpow(int a, long long b) {

    int ret = 1; while (b > 0) {

        if (b & 1) ret = 1LL * ret * a % mod; a = 1LL * a * a % mod; b >>= 1;

    } return ret;

}

int modinv(int a) { return modpow(a, mod - 2); }

#define def 0

template<class V, int NV> struct SegTree {

    V comp(V l, V r) { return add(l,r); };

    vector<V> val; SegTree() { val = vector<V>(NV * 2, def); }

    V get(int l, int r) { //[l,r]

        if (l > r) return def;

        l += NV; r += NV + 1; V ret = def;

        while (l < r) { if (l & 1) ret = comp(ret, val[l++]); if (r & 1) ret = comp(ret, val[--r]); l /= 2; r /= 2; }

        return ret;

    }

    V get(int x) { return val[x + NV]; }

    void update(int i, V v) { i += NV; val[i] = v; while (i>1) i >>= 1, val[i] = comp(val[i * 2], val[i * 2 + 1]); }

};

/*---------------------------------------------------------------------------------------------------

　　　　　　　　　　　 ∧＿∧  

　　　　　 ∧＿∧ 　（´<_｀ ）　 Welcome to My Coding Space!

　　　　 （ ´_ゝ`）　/　 ⌒i     

　　　　／　　　＼　 　  |　|     

　　　 /　　 /￣￣￣￣/　　|  

　 ＿_(__ﾆつ/　    ＿/ .| .|＿＿＿＿  

　 　　　＼/＿＿＿＿/　（u　⊃  

---------------------------------------------------------------------------------------------------*/









int N;

pair<int, int> P[201010];

//---------------------------------------------------------------------------------------------------

void zaatsu() {

    vector<int> dic;

    rep(i, 0, N) dic.push_back(P[i].second);

    sort(dic.begin(), dic.end());

    dic.erase(unique(dic.begin(), dic.end()), dic.end());



    rep(i, 0, N) P[i].second = lower_bound(dic.begin(), dic.end(), P[i].second) - dic.begin() + 1;



    /*rep(i, 0, N) printf("%d %d\n", L[i], R[i]);

    printf("<%d>\n", M);*/

}

//---------------------------------------------------------------------------------------------------

#define INF INT_MAX/2

#define rrep(i,a,b) for(int i=a;i>=b;i--)

int L[201010], R[201010];

void change() {

    sort(P, P + N);



    rep(i, 0, N) {

        int pre = 0;

        if (0 < i) pre = R[i - 1];

        R[i] = max(pre, P[i].second);

    }

    

    L[N] = INF;

    rrep(i, N - 1, 0) {

        int pre = L[i + 1];

        L[i] = min(pre, P[i].second);

    }



    //rep(i, 0, N) printf("%d %d\n", L[i], R[i]);

}

//---------------------------------------------------------------------------------------------------

// Solution1 O(N^2)

/*int dp[2020][2020];

int dodp() {

    dp[0][0] = 1;

    rep(i, 0, N) {

        rep(j, 0, M + 1) dp[i + 1][j] = dp[i][j];



        int sm = 0;

        rep(j, L[i] - 1, R[i] + 1) sm = add(sm, dp[i][j]);

        dp[i + 1][R[i]] = add(dp[i + 1][R[i]], sm);

    }



    return dp[N][N];

}*/

//---------------------------------------------------------------------------------------------------

// Solution2 O(N^2)

/*int dp[201010];

int dodp() {

    dp[0] = 1;

    rep(i, 0, N) {

        int sm = 0;

        rep(j, L[i] - 1, R[i] + 1) sm = add(sm, dp[j]);

        dp[R[i]] = add(dp[R[i]], sm);

    }



    return dp[N];

}*/

//---------------------------------------------------------------------------------------------------

// Solution3 O(NlogN)

/*SegTree<int, 1 << 18> dp;

int dodp() {

    dp.update(0, 1);

    rep(i, 0, N) {

        int sm = dp.get(L[i] - 1, R[i]);

        dp.update(R[i], add(dp.get(R[i]), sm));

    }



    return dp.get(N);

}*/

//---------------------------------------------------------------------------------------------------

// Solution4 O(N)

int dp[201010];

int sm[201010];

int dodp() {

    dp[0] = 1;

    sm[0] = 1;

    int r = 0;



    rep(i, 0, N) {

        while (r < R[i] - 1) {

            r++;

            sm[r] = add(sm[r - 1], dp[r]);

        }



        int s = sm[R[i] - 1];

        if (0 <= L[i] - 2) s = sub(s, sm[L[i] - 2]);

        s = add(s, dp[R[i]]);

        dp[R[i]] = add(dp[R[i]], s);

    }



    return dp[N];

}

//---------------------------------------------------------------------------------------------------

void _main() {

    cin >> N;

    rep(i, 0, N) cin >> P[i].first >> P[i].second;



    zaatsu();

    change();



    cout << dodp() << endl;

}