// Slower C++ solution:
#include <iostream>

#include <cstdio>

#include <string>

#include <vector>

#include <algorithm>

#include <functional>

#include <iomanip>

#include <stdlib.h>

#include <string.h>

#include <cstring>

#include <cmath>

#include <map>

#include <queue>

#include <deque>

#include <stack>

#include <set>

#include <stdio.h>

#include <tuple>

#include <bitset>

#include <cfloat>

#include <fstream>

#include <limits.h>

#include <list>

#include <math.h>

#include <random>

#include <unordered_map>

#include <unordered_set>

#include <bitset>

#include <numeric>

#include <utility>

#include <sstream>

#include <cassert>

template<class T> inline bool chmin(T& a, T b) {

	if (a > b) {

		a = b;

		return true;

	}

	return false;

}

template<class T> inline bool chmax(T& a, T b) {

	if (a < b) {

		a = b;

		return true;

	}

	return false;

}

using namespace std;

#define int long long

#define ll long long

#define rep(i, n) for (ll i = 0; i < (n); i++)

#define P pair<ll, ll>

#define sz(x) (ll)x.size()

#define ALL(x) (x).begin(),(x).end()

#define ALLR(x) (x).rbegin(),(x).rend()

#define VE vector<ll>

#define COUT(x) cout<<(x)<<endl

#define MA map<ll,ll>

#define SE set<ll>

#define PQ priority_queue<ll>

#define PQR priority_queue<ll,VE,greater<ll>>

#define YES(n) cout << ((n) ? "YES" : "NO"  ) << endl

#define Yes(n) cout << ((n) ? "Yes" : "No"  ) << endl

#define EPS (1e-14)

#define pb push_back

long long MOD = 1000000007;

//const long long MOD = 998244353;

const long long INF = 1LL << 60;

const double PI = acos(-1.0);

struct mint {

	ll x; // typedef long long ll;

	mint(ll x = 0) :x((x%MOD + MOD) % MOD) {}

	mint operator-() const { return mint(-x); }

	mint& operator+=(const mint a) {

		if ((x += a.x) >= MOD) x -= MOD;

		return *this;

	}

	mint& operator-=(const mint a) {

		if ((x += MOD - a.x) >= MOD) x -= MOD;

		return *this;

	}

	mint& operator*=(const mint a) {

		(x *= a.x) %= MOD;

		return *this;

	}

	mint operator+(const mint a) const {

		mint res(*this);

		return res += a;

	}

	mint operator-(const mint a) const {

		mint res(*this);

		return res -= a;

	}

	mint operator*(const mint a) const {

		mint res(*this);

		return res *= a;

	}

	mint pow(ll t) const {

		if (!t) return 1;

		mint a = pow(t >> 1);

		a *= a;

		if (t & 1) a *= *this;

		return a;

	}

	// for prime MOD

	mint inv() const {

		return pow(MOD - 2);

	}

	mint& operator/=(const mint a) {

		return (*this) *= a.inv();

	}

	mint operator/(const mint a) const {

		mint res(*this);

		return res /= a;

	}

};

istream& operator>>(istream& is, const mint& a) { return is >> a.x; }

ostream& operator<<(ostream& os, const mint& a) { return os << a.x; }

struct combination {

	vector<mint> fact, ifact;

	combination(int n) :fact(n + 1), ifact(n + 1) {

		assert(n < MOD);

		fact[0] = 1;

		for (int i = 1; i <= n; ++i) fact[i] = fact[i - 1] * i;

		ifact[n] = fact[n].inv();

		for (int i = n; i >= 1; --i) ifact[i - 1] = ifact[i] * i;

	}

	mint operator()(int n, int k) {

		if (k < 0 || k > n) return 0;

		return fact[n] * ifact[k] * ifact[n - k];

	}

}com(10);

struct Sieve {

	int n;

	vector<int> f, primes;

	

	Sieve(int n = 1) :n(n), f(n + 1) {

		f[0] = f[1] = -1;

		for (ll i = 2; i <= n; ++i) {

			if (f[i]) continue;

			primes.push_back(i);

			f[i] = i;

			for (ll j = i * i; j <= n; j += i) {

				if (!f[j]) f[j] = i;

			}

		}

	}

	

	bool isPrime(int x) { return f[x] == x; }

	

	vector<int> factorList(int x) {

		vector<int> res;

		while (x != 1) {

			res.push_back(f[x]);

			x /= f[x];

		}

		return res;

	}

	

	vector<P> factor(int x) {

		vector<int> fl = factorList(x);

		if (fl.size() == 0) return {};

		vector<P> res(1, P(fl[0], 0));

		for (int p : fl) {

			if (res.back().first == p) {

				res.back().second++;

			}

			else {

				res.emplace_back(p, 1);

			}

		}

		return res;

	}

};

template<class t> t gcd(t a, t b) { return b != 0 ? gcd(b, a % b) : a; }

ll lcm(ll a, ll b) {

	ll g = gcd(a, b);

	return a / g * b;

}

bool prime(ll n) {

	for (ll i = 2; i <= sqrt(n); i++) {

		if (n%i == 0)return false;

	}

	return n != 1;

}

map<ll, ll> prime_factor(ll n) {

	map<ll, ll> ret;

	for (ll i = 2; i * i <= n; i++) {

		while (n % i == 0) {

			ret[i]++;

			n /= i;

		}

	}

	if (n != 1) ret[n] = 1;

	return ret;

}

ll modinv(ll a, ll m) {

	ll b = m, u = 1, v = 0;

	while (b) {

		ll t = a / b;

		a -= t * b; swap(a, b);

		u -= t * v; swap(u, v);

	}

	u %= m;

	if (u < 0) u += m;

	return u;

}

vector<pair<char, int>> RunLength(string s) {

	if (s.size() == 0)return {};

	vector<pair<char, int>>res(1, pair<char, int>(s[0], 0));

	for (char p : s) {

		if (res.back().first == p) {

			res.back().second++;

		}

		else {

			res.emplace_back(p, 1);

		}

	}

	return res;

}

// Digit Count

int GetDigit(int num) {

	return log10(num) + 1;

}

// bit calculation[how many "1"] (= __builtin_popcount())

int bit_count(int n) {

	int cnt = 0;

	while (n > 0) {

		if (n % 2 == 1)cnt++;

		n /= 2;

	}

	return cnt;

}



const ll dx[4] = { 1,0,-1,0 };

const ll dy[4] = { 0,1,0,-1 };

struct edge { ll to, cost; };

typedef long double ld;

using Graph = vector<VE>;

int dp[1000010];

signed main() {

	ios_base::sync_with_stdio(0);

	cin.tie(0);

	cout.tie(0);

	//cout << fixed << setprecision(15);

	int h, n; cin >> h >> n;

	VE a(n), b(n);

	rep(i, n) {

		cin >> a[i] >> b[i];

	}

	rep(i, 1000010)dp[i] = INF;

	rep(i, n) {

		for (int j = 0; j <= a[i]; j++) {

			chmin(dp[j], b[i]);

		}

		for (int j = a[i]; j <= h; j++) {

			chmin(dp[j], dp[j - a[i]] + b[i]);

		}

	}

	cout << dp[h] << endl;

	int ans = INF;

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 4.787154175389643x):
#include "bits/stdc++.h"

using namespace std;



// コンテスト中のみ

//#define int long long



#define ll long long

#define rep(i,n) for(int i = 0; i < (n); i++)

#define P pair<ll,ll>

int MOD = 1000000007;

const ll INF = 1LL << 60;



ll H, N;

ll A[1010], B[1010];

ll dp[100010];

signed main() {

	ios::sync_with_stdio(false);

	cin.tie(nullptr);

	cin >> H >> N;

	rep(i, N)cin >> A[i] >> B[i];

	rep(i, 100010)dp[i] = INF;

	rep(i, N) {

		rep(j, A[i]) {

			dp[j] = min(dp[j], B[i]);

		}

		for (int j = A[i]; j <= H; j++) {

			dp[j] = min(dp[j], dp[j - A[i]] + B[i]);

		}

	}

	cout << dp[H - 1] << endl;

	return 0;

}