Problem: Count the number of strings S that satisfy the following constraints, modulo 10^9 + 7.

* The length of S is exactly N.
* S consists of digits (`0`...`9`).
* You are given Q intervals. For each i (1 \leq i \leq Q), the integer represented by S[l_i \ldots r_i] (the substring of S between the l_i-th (1-based) character and the r_i-th character, inclusive) must be a multiple of 9.



Here, the string S and its substrings may have leading zeroes. For example, `002019` represents the integer 2019.

Constraints

* 1 \leq N \leq 10^9
* 1 \leq Q \leq 15
* 1 \leq l_i \leq r_i \leq N

Input

Input is given from Standard Input in the following format:


N
Q
l_1 r_1
:
l_Q r_Q


Output

Print the number of strings that satisfy the conditions, modulo 10^9 + 7.

Examples

Input

4
2
1 2
2 4


Output

136


Input

6
3
2 5
3 5
1 3


Output

2720


Input

20
10
2 15
5 6
1 12
7 9
2 17
5 15
2 4
16 17
2 12
8 17


Output

862268030

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
//#pragma GCC optimize("Ofast")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")

#define ms(s, n) memset(s, n, sizeof(s))
#define FOR(i, a, b) for (int i = (a); i < (b); ++i)
#define FORd(i, a, b) for (int i = (a) - 1; i >= (b); --i)
#define FORall(it, a) for (__typeof((a).begin()) it = (a).begin(); it != (a).end(); it++)
#define sz(a) int((a).size())
#define present(t, x) (t.find(x) != t.end())
#define all(a) (a).begin(), (a).end()
#define uni(a) (a).erase(unique(all(a)), (a).end())
#define pb push_back
#define pf push_front
#define mp make_pair
#define fi first
#define se second
#define prec(n) fixed<<setprecision(n)
#define bit(n, i) (((n) >> (i)) & 1)
#define bitcount(n) __builtin_popcountll(n)
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef pair<int, int> pi;
typedef vector<int> vi;
typedef vector<pi> vii;
const int MOD = (int) 1e9 + 7;
const int FFTMOD = 119 << 23 | 1;
const int INF = (int) 1e9 + 23111992;
const ll LINF = (ll) 1e18 + 23111992;
const ld PI = acos((ld) -1);
const ld EPS = 1e-9;
inline ll gcd(ll a, ll b) {ll r; while (b) {r = a % b; a = b; b = r;} return a;}
inline ll lcm(ll a, ll b) {return a / gcd(a, b) * b;}
inline ll fpow(ll n, ll k, int p = MOD) {ll r = 1; for (; k; k >>= 1) {if (k & 1) r = r * n % p; n = n * n % p;} return r;}
template<class T> inline int chkmin(T& a, const T& val) {return val < a ? a = val, 1 : 0;}
template<class T> inline int chkmax(T& a, const T& val) {return a < val ? a = val, 1 : 0;}
inline ull isqrt(ull k) {ull r = sqrt(k) + 1; while (r * r > k) r--; return r;}
inline ll icbrt(ll k) {ll r = cbrt(k) + 1; while (r * r * r > k) r--; return r;}
inline void addmod(int& a, int val, int p = MOD) {if ((a = (a + val)) >= p) a -= p;}
inline void submod(int& a, int val, int p = MOD) {if ((a = (a - val)) < 0) a += p;}
inline int mult(int a, int b, int p = MOD) {return (ll) a * b % p;}
inline int inv(int a, int p = MOD) {return fpow(a, p - 2, p);}
inline int sign(ld x) {return x < -EPS ? -1 : x > +EPS;}
inline int sign(ld x, ld y) {return sign(x - y);}
mt19937 mt(chrono::high_resolution_clock::now().time_since_epoch().count());
inline int myrand() {return abs((int) mt());}
#define db(x) cerr << #x << " = " << (x) << " ";
#define endln cerr << "\n";

int calc(int n, int k) {
    int res = fpow(10, n);
    submod(res, 1);
    res = mult(res, 111111112);
    if (!k) addmod(res, 1);
    return res;
}

void chemthan() {
    vector<vi> c(20, vi(20));
    FOR(i, 0, 20) c[0][i] = 1;
    FOR(i, 1, 20) FOR(j, 1, 20) {
        c[i][j] = c[i][j - 1] + c[i - 1][j - 1];
        c[i][j] %= MOD;
    }
    int n, q; cin >> n >> q;
    vii vals(q);
    for (pi& e : vals) cin >> e.fi >> e.se;
    vi dc;
    for (pi& e : vals) e.fi--, dc.pb(e.fi), dc.pb(e.se);
    sort(all(dc)), uni(dc);
    vi f(sz(dc), -1), g(sz(dc), -1);
    for (pi& e : vals) e.fi = lower_bound(all(dc), e.fi) - dc.begin();
    for (pi& e : vals) e.se = lower_bound(all(dc), e.se) - dc.begin();
    FOR(i, 0, sz(vals)) {
        f[vals[i].fi] = i;
        g[vals[i].se] = i;
    }
    vi can(1 << sz(vals), 1);
    FOR(msk, 0, 1 << sz(vals)) {
        FOR(i, 0, sz(vals)) FOR(j, i + 1, sz(vals)) if (bit(msk, i) ^ bit(msk, j)) {
            if (vals[i].fi == vals[j].se || vals[i].se == vals[j].fi || vals[i].fi == vals[j].fi || vals[i].se == vals[j].se) {
                can[msk] = 0;
            }
        }
    }
    vector<vi> dp(1 << sz(vals), vi(10));
    FOR(msk, 1, 1 << sz(vals)) if (can[msk]) {
        dp[msk][1] = 1;
        FOR(i, 0, sz(dc) - 1) {
            int u = dc[i];
            int v = dc[i + 1];
            if (bit(msk, max(f[i], g[i])) && bit(msk, max(f[i + 1], g[i + 1]))) {
                dp[msk][1] = mult(dp[msk][1], calc(v - u, 0));
            }
            else if (bit(msk, max(f[i + 1], g[i + 1]))) {
                dp[msk][1] = mult(dp[msk][1], calc(v - u, 1));
            }
        }
    }
    FOR(msk, 1, 1 << sz(vals)) {
        for (int sub = msk; ; sub = sub - 1 & msk) {
            int psub = sub ^ msk;
            int u = bitcount(sub);
            int v = bitcount(psub);
            if (u && v && (sub & -sub) == (msk & -msk)) {
                FOR(i, 1, v + 1) if (i + 1 < 10) {
                    addmod(dp[msk][i + 1], mult(dp[sub][1], dp[psub][i]));
                }
            }
            if (!sub) break;
        }
    }
    vi fac(10);
    fac[0] = 1; FOR(i, 1, 10) fac[i] = mult(fac[i - 1], i);
    int res = 0;
    FOR(i, 1, 9 + 1) {
        int coef = 0;
        if (!dc[0]) {
            coef = mult(c[i - 1][8], fac[i - 1]);
        }
        else {
            FOR(d, 0, 9) {
                if (!d) {
                    addmod(coef, mult(calc(dc[0], 0), mult(c[i - 1][8], fac[i - 1])));
                }
                else {
                    addmod(coef, mult(calc(dc[0], 1), mult(c[i - 1][8], fac[i - 1])));
                }
            }
        }
        addmod(res, mult(dp.back()[i], coef));
    }
    res = mult(res, fpow(10, n - dc.back()));
    cout << res << "\n";
}

int main(int argc, char* argv[]) {
    ios_base::sync_with_stdio(0), cin.tie(0);
    if (argc > 1) {
        assert(freopen(argv[1], "r", stdin));
    }
    if (argc > 2) {
        assert(freopen(argv[2], "wb", stdout));
    }
    chemthan();
    cerr << "\nTime elapsed: " << 1000 * clock() / CLOCKS_PER_SEC << "ms\n";
    return 0;
}