This version of the problem differs from the next one only in the constraint on n.

Note that the memory limit in this problem is lower than in others.

You have a vertical strip with n cells, numbered consecutively from 1 to n from top to bottom.

You also have a token that is initially placed in cell n. You will move the token up until it arrives at cell 1.

Let the token be in cell x > 1 at some moment. One shift of the token can have either of the following kinds: 

  * Subtraction: you choose an integer y between 1 and x-1, inclusive, and move the token from cell x to cell x - y. 
  * Floored division: you choose an integer z between 2 and x, inclusive, and move the token from cell x to cell ⌊ x/z ⌋ (x divided by z rounded down). 



Find the number of ways to move the token from cell n to cell 1 using one or more shifts, and print it modulo m. Note that if there are several ways to move the token from one cell to another in one shift, all these ways are considered distinct (check example explanation for a better understanding).

Input

The only line contains two integers n and m (2 ≤ n ≤ 2 ⋅ 10^5; 10^8 < m < 10^9; m is a prime number) — the length of the strip and the modulo.

Output

Print the number of ways to move the token from cell n to cell 1, modulo m.

Examples

Input


3 998244353


Output


5


Input


5 998244353


Output


25


Input


42 998244353


Output


793019428

Note

In the first test, there are three ways to move the token from cell 3 to cell 1 in one shift: using subtraction of y = 2, or using division by z = 2 or z = 3.

There are also two ways to move the token from cell 3 to cell 1 via cell 2: first subtract y = 1, and then either subtract y = 1 again or divide by z = 2.

Therefore, there are five ways in total.


#include <bits/stdc++.h>
using namespace std;

#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;

//========================Debug======================================

void __print(int x) { cerr << x; }
void __print(long x) { cerr << x; }
void __print(long long x) { cerr << x; }
void __print(unsigned x) { cerr << x; }
void __print(unsigned long x) { cerr << x; }
void __print(unsigned long long x) { cerr << x; }
void __print(float x) { cerr << x; }
void __print(double x) { cerr << x; }
void __print(long double x) { cerr << x; }
void __print(char x) { cerr << '\'' << x << '\''; }
void __print(const char *x) { cerr << '\"' << x << '\"'; }
void __print(const string &x) { cerr << '\"' << x << '\"'; }
void __print(bool x) { cerr << (x ? "true" : "false"); }

template <typename T, typename V>
void __print(const pair<T, V> &x)
{
    cerr << '{';
    __print(x.first);
    cerr << ',';
    __print(x.second);
    cerr << '}';
}
template <typename T>
void __print(const T &x)
{
    int f = 0;
    cerr << '{';
    for (auto &i : x)
        cerr << (f++ ? "," : ""), __print(i);
    cerr << "}";
}
void _print() { cerr << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v)
{
    __print(t);
    if (sizeof...(v))
        cerr << ", ";
    _print(v...);
}
#ifdef local
#define debug(x...)               \
    cerr << "[" << #x << "] = ["; \
    _print(x)
#else
#define debug(x...)
#endif

//========================TypeDefs===================================

typedef long long int ll;
typedef unsigned long long int ull;
typedef long double ldb;

typedef pair<ll, ll> pll;
typedef pair<int, int> pii;
typedef pair<int, ll> pil;

//=========================MACROS====================================

#define pb push_back
#define popb pop_back()
#define pf push_front
#define popf pop_front()
#define si size()
#define be begin()
#define en end()
#define all(v) v.be, v.en
#define mp make_pair
#define mt make_tuple
#define acc(v) accumulate(all(v), 0)
#define F first
#define S second

#define forz(i, n) for (int i = 0; i < n; i++)
#define fore(i, m, n) for (int i = m; i <= n; i++)
#define rforz(i, n) for (int i = n - 1; i >= 0; i--)
#define rfore(i, m, n) for (int i = n; i >= m; i--)

#define deci(n) fixed << setprecision(n)
#define high(n) __builtin_popcount(n)
#define highll(n) __builtin_popcountll(n)
#define parity(n) __builtin_parity(n)
#define clz(n) __builtin_clz(n)
#define clzll(n) __builtin_clzll(n)
#define ctz(n) __builtin_ctz(n)

#define bset(a, p) ((a) | (1ll << (p)))
#define bchk(a, p) ((a) & (1ll << (p)))
#define bxor(a, p) ((a) ^ (1ll << (p)));
#define brem(a, p) (bchk(a, p) ? (bxor(a, p)) : (a))

#define lb lower_bound
#define ub upper_bound
#define er equal_range

#define findnot find_first_not_of

#define maxe *max_element
#define mine *min_element

#define mod 1000000007
#define mod2 998244353
#define PI 3.1415926535897932384
#define INF LLONG_MAX

#define gcd __gcd
#define kira ios::sync_with_stdio(0), cin.tie(0), cout.tie(0)

#define endl "\n"
#define p0(a) cout << a << " "
#define p1(a) cout << a << endl
#define p2(a, b) cout << a << " " << b << endl
#define p3(a, b, c) cout << a << " " << b << " " << c << endl
#define p4(a, b, c, d) cout << a << " " << b << " " << c << " " << d << endl

#define oset tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update>
#define osetlli tree<ll, null_type, less<ll>, rb_tree_tag, tree_order_statistics_node_update>

#define ofk order_of_key
#define fbo find_by_order
//member functions :
//1. order_of_key(k) : number of elements sbtriectly lesser than k
//2. find_by_order(k) : k-th element in the set

//==============================FUNCTIONS===========================================

auto clk = clock();
mt19937_64 rang(chrono::high_resolution_clock::now().time_since_epoch().count());
void run_time()
{
#ifdef local
    cout << endl;
    cout << "Time elapsed: " << (double)(clock() - clk) / CLOCKS_PER_SEC << endl;
#endif
    return;
}

inline ll power(ll x, ll y, ll p = mod)
{
    ll res = 1;
    x = x % p;
    while (y > 0)
    {
        if (y & 1)
            res = (res * x) % p;
        y = y >> 1;
        x = (x * x) % p;
    }
    return res;
}

inline ll modadd(ll a, ll b, ll m = mod)
{
    a += b;
    if (a >= m)
        a -= m;
    return a;
}

inline ll modmul(ll a, ll b, ll m = mod)
{
    return ((a % m) * (b % m)) % m;
}

inline ll modi(ll a, ll m = mod) { return power(a, m - 2, m); }

//================================CODE=============================================


void solve(){
    ll n,m;
    cin >>n>>m;
    ll dp[n+1]={0};
    dp[1]=1;
    dp[2]=2;
    for(ll i=2;i*2<=n;i++){
        dp[i*2]+=dp[2]-dp[1];
    }
    for(ll i=3;i<=n;i++){
        dp[i]=(dp[i]+(dp[i-1]+dp[i-1]))%m;
        dp[i]=(dp[i]+dp[1])%m;
        for(ll j=2;j*i<=n;j++){
            dp[j*i]=(dp[j*i]+(dp[i]-dp[i-1]))%m;
        }
    }
    cout << (dp[n])%m;
}


int main()
{
    kira;
    ll t;
   // cin >>t;
    t=1;
    while(t--)solve();

    run_time();
    return 0;
}