#include <bits/stdc++.h>
using namespace std;
typedef pair<int,int> pii;
typedef vector<int> vi;
typedef long long ll;
typedef pair<ll,ll> pll;
typedef tuple<int,int,int> ti;
typedef unsigned long long ull;
typedef long double ld;
typedef pair<ld,ld> pld;
#define pb push_back
#define popb pop_back()
#define pf push_front
#define popf pop_front
#define ff first
#define ss second
#define MOD (int)(1e8)
#define INF (ll) (1e18)
ll gcd(ll a , ll b) {return b ? gcd(b , a % b) : a ;}
ll lcm(ll a , ll b) {return (a * b) / gcd(a , b);}
ld pointdist(ld x, ld y, ld point) { return ((x-point)*(y-point))/abs(x-y); }
ld dist(ld x, ld y, ld a, ld b){ return sqrt((x-a)*(x-a) + (y-b)*(y-b)); }
const int nx[4] = {0, 0, 1, -1}, ny[4] = {1, -1, 0, 0}; //East, West, South, North
const int xcav[8] = {1,2,2,1,-1,-2,-2,-1};
const int ycav[8] = {-2,-1,1,2,2,1,-1,-2};
const int crapx[4] = {0,2,0,-2};
const int crapy[4] = {-2,0,2,0};
////////////******SOLUTION******\\\\\\\\\\\
const int MAX_N = 2e5 + 4;
int n, k, r;
vector<int> dna;
int m[MAX_N];
bool f(int x)
{
int seen[k];
memset(seen,0,sizeof(seen));
int cnt = 0;
for(int i = 0; i <x; i ++)
{
seen[dna[i]] ++;
if(seen[dna[i]] == m[dna[i]])
cnt ++;
}
if(cnt == r)
return true;
for(int i = x; i <n; i ++)
{
if(m[dna[i-x]] != -1)
{
seen[dna[i-x]] --;
if(seen[dna[i-x]] == m[dna[i-x]] - 1)
cnt --;
}
if(m[dna[i]] != -1)
{
seen[dna[i]] ++;
if(seen[dna[i]] == m[dna[i]])
cnt ++;
}
if(cnt == r)
return true;
}
return false;
}
int main()
{
ios::sync_with_stdio(false);
cin.tie(0); cout.tie(0);
cin >> n >> k >> r;
memset(m,-1,sizeof(m));
for(int i = 0; i <n; i ++ )
{
int a; cin >> a;
dna.pb(a);
}
for(int i = 0; i < r; i ++)
{
int a, b;
cin >> a >> b;
m[a] = b;
}
int ans = -1;
int debut = 1;
int fin = n;
while(debut <= fin)
{
int mid = (debut + fin)/2;
if(f(mid))
{
ans = mid;
fin = mid - 1;
}
else
debut = mid + 1;
}
if(ans == -1)
cout << "impossible\n";
else
cout << ans << '\n';
}
/* Case that bugs my program:
5280
11
54 19 70 69 33 27 91 77 30 47 6
*/
/*
Identify problem diagram: Brute force, Greedy, Dynamic Programming, Divide and Conquer
Reformulate the problem into something more theoretical
Search for multiple approaches: select the seemingly optimal one
Remember that you're the king of CP
Change your approach
Imagine Corner cases before submitting
Don't spend too much time on the problem: move on !
*/
